Chapter 6 Calculations

$R_1 = 2\text{ }\Omega,\quad I_1 = 0.6\text{ A}$
$R_2 = 7\text{ }\Omega,\quad I_2 = 0.2\text{ A}$
$E = ?,\quad r = ?$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ I_1 &= \frac{E}{R_1 + r} \\ E &= I_1 (R_1 + r) \\ &= 0.6 (2 + r) \text{ } -------- (1) \\ I_2 &= \frac{E}{R_2 + r} \\ E &= I_2 (R_2 + r) \\ &= 0.2 (7 + r) \text{ } -------- (2) \end{aligned}$

From Eq (1) and (2)
$\begin{aligned}[t] 0.6 (2 + r) &= 0.2 (7 + r) \\ \frac{0.6}{0.2} (2 + r) &= 7 + r \\ 3 (2 + r) &= 7 + r \\ 6 + 3r &= 7 + r \\ 2r &= 1 \\ r &= \mathbf{0.5\text{ }\Omega} \end{aligned}$

$r = 0.5\text{ }\Omega$ in eq (1)
$\begin{aligned}[t] E &= 0.6 (2 + 0.5) \\ &= \mathbf{1.5\text{ V}} \end{aligned}$

$R_1 = 3\text{ }\Omega,\quad R_2 = R_3 = 2\text{ }\Omega$
$E = 20\text{ V},\quad r = 0\text{ }\Omega$

(i) When the switch is open,
$R_1$ and $R_2$ are connected in series,
$\begin{aligned}[t] R &= R_1 + R_2 \\ &= 3 + 2 = 5\text{ }\Omega \end{aligned}$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ &= \frac{20}{5 + 0} \\ &= \mathbf{4\text{ A}} \end{aligned}$

$\therefore$ When the switch is open, the reading of the ammeter A is $4\text{ A}$.

(ii) When the switch is closed,
$R_2$ and $R_3$ are connected in parallel,
$\begin{aligned}[t] \frac{1}{R_p} &= \frac{1}{R_2} + \frac{1}{R_3} \\ &= \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1 \\ R_p &= 1\text{ }\Omega \end{aligned}$

$R_p$ and $R_1$ are connected in series,
$\begin{aligned}[t] R &= R_1 + R_p \\ &= 3 + 1 = 4\text{ }\Omega \end{aligned}$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ &= \frac{20}{4 + 0} = \frac{20}{4} = \mathbf{5\text{ A}} \end{aligned}$

$\therefore$ When the switch is closed, the reading of the ammeter A is $5\text{ A}$.

$R_1 = 4\text{ }\Omega,\quad R_2 = 6\text{ }\Omega,\quad R_3 = 1\text{ }\Omega$
$E = 12\text{ V},\quad r = 0.6\text{ }\Omega$
$I_1 = ?,\quad I_2 = ?,\quad I_3 = ?$
$V_1 = ?,\quad V_2 = ?,\quad V_3 = ?$

$R_1$ and $R_2$ are connected in parallel,
$\begin{aligned}[t] \frac{1}{R_p} &= \frac{1}{R_1} + \frac{1}{R_2} \\ &= \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \\ R_p &= \frac{12}{5} = 2.4\text{ }\Omega \end{aligned}$

$R_p$ and $R_3$ are connected in series,
$\begin{aligned}[t] R &= R_p + R_3 \\ &= 2.4 + 1 = 3.4\text{ }\Omega \end{aligned}$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ &= \frac{12}{3.4 + 0.6} = \frac{12}{4.0} = \mathbf{3\text{ A}} \end{aligned}$

$I = I_3 = I_p = \mathbf{3\text{ A}}$ (in series)

By Ohm's law, $V = IR$
$\begin{aligned}[t] V_3 &= I_3 R_3 = 3 \times 1 = \mathbf{3\text{ V}} \\ V_p &= I_p R_p = 3 \times 2.4 = \mathbf{7.2\text{ V}} \\ V_p &= V_1 = V_2 = \mathbf{7.2\text{ V}} \text{ (in parallel)} \end{aligned}$

Since $V = IR$, $I = \frac{V}{R}$
$\begin{aligned}[t] I_1 &= \frac{V_1}{R_1} = \frac{7.2}{4} = \mathbf{1.8\text{ A}} \\ I_2 &= \frac{V_2}{R_2} = \frac{7.2}{6} = \mathbf{1.2\text{ A}} \end{aligned}$

$E_1 = E_2 = 6\text{ V},\quad r_1 = r_2 = 0.5\text{ }\Omega$
$R = 1\text{ }\Omega,\quad I = ?$

(i) When the batteries are connected in series,
In series aiding,
Resultant emf, $E = E_1 + E_2 = 6 + 6 = 12\text{ V}$
Resultant $r = r_1 + r_2 = 0.5 + 0.5 = 1\text{ }\Omega$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ &= \frac{12}{1 + 1} = \mathbf{6\text{ A}} \end{aligned}$

In series opposing,
Resultant emf, $E = E_1 - E_2 = 6 - 6 = 0\text{ V}$
Resultant $r = r_1 + r_2 = 0.5 + 0.5 = 1\text{ }\Omega$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ &= \frac{0}{1 + 1} = \mathbf{0\text{ A}} \end{aligned}$

(ii) When the batteries are connected in Parallel,
Resultant emf, $E = E_1 = E_2 = 6\text{ V}$
Resultant $r = \frac{r_1}{2} = \frac{0.5}{2} = 0.25\text{ }\Omega$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ &= \frac{6}{1 + 0.25} = \mathbf{4.8\text{ A}} \end{aligned}$

$E_1 = 6\text{ V},\quad r_1 = 0.5\text{ }\Omega,\quad R = 22\text{ }\Omega,\quad I = 1\text{ A}$
Number of batteries, $n = ?$

By Ohm's law, $V = I R = 1 \times 22 = 22\text{ V}$
Since $V > E$, the batteries must be connected in series aiding.

In series aiding,
Resultant emf, $E = E_1 + E_2 ...... + E_n = 6 + 6 + ...... + 6 = 6n$
Resultant $r = r_1 + r_2 ...... + r_n = 0.5 + 0.5 + ...... + 0.5 = 0.5n$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ &= \frac{6n}{R + 0.5n} \\ I (R + 0.5n) &= 6n \\ 1 (22 + 0.5n) &= 6n \\ 22 + 0.5n &= 6n \\ 22 &= 5.5n \\ n &= \frac{22}{5.5} = \mathbf{4} \end{aligned}$

$\therefore$ $4$ batteries are necessary.

$E = 12\text{ V},\quad r = 0\text{ }\Omega,\quad I = 3\text{ A}$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ 3 &= \frac{12}{R + 0} \\ 3 R &= 12 \\ R &= \mathbf{4\text{ }\Omega} \end{aligned}$

$E_1 = 12\text{ V},\quad r_1 = 0\text{ }\Omega,\quad R = 4\text{ }\Omega$
$E_2 = 6\text{ V},\quad r_2 = ?,\quad I = 3\text{ A}$

In series aiding,
Resultant emf, $E = E_1 + E_2 = 12 + 6 = 18\text{ V}$
Resultant $r = r_1 + r_2 = 0 + r_2 = r_2$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ 3 &= \frac{18}{4 + r_2} \\ 12 + 3r_2 &= 18 \\ 3r_2 &= 6 \\ r_2 &= \mathbf{2\text{ }\Omega} \end{aligned}$

$\therefore$ The internal resistance of the second battery is $2\text{ }\Omega$.

$E_1 = E_2 = 6\text{ V},\quad r_1 = r_2,\quad R = 5\text{ }\Omega,\quad I = 2\text{ A}$

In series aiding,
Resultant emf, $E = E_1 + E_2 = 6 + 6 = 12\text{ V}$
Resultant $r = r_1 + r_2 = r_1 + r_1 = 2r_1$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} \\ 2 &= \frac{12}{5 + 2r_1} \\ 10 + 4r_1 &= 12 \\ 4r_1 &= 2 \\ r_1 &= 0.5\text{ }\Omega \end{aligned}$

$r_1 = r_2 = r = 0.5\text{ }\Omega$

In parallel,
Resultant emf, $E = E_1 = E_2 = 6\text{ V}$
Resultant $r = \frac{r_1}{2} = \frac{0.5}{2} = 0.25\text{ }\Omega$
$I = 1.5\text{ A},\quad R' = ?$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R' + r} \\ 1.5 &= \frac{6}{R' + 0.25} \\ 1.5 (R' + 0.25) &= 6 \\ R' + 0.25 &= \frac{6}{1.5} \\ R' + 0.25 &= 4 \\ R' &= 4 - 0.25 \\ &= \mathbf{3.75\text{ }\Omega} \end{aligned}$

$I = 2\text{ A},\quad R = 50\text{ }\Omega$
$t = 30\text{ min} = 30\text{ min} \times \frac{1\text{ h}}{60\text{ min}} = \frac{1}{2}\text{ h}$

$\begin{aligned}[t] W &= I^2 R t \\ &= 2^2 \times 50 \times \frac{1}{2} \\ &= 100\text{ Wh} \\ &= 100\text{ Wh} \times \frac{1\text{ kWh}}{1000\text{ Wh}} \\ &= \mathbf{0.1\text{ kWh}} \end{aligned}$

$R = 60\text{ }\Omega,\quad V = 240\text{ V}$
$t = 45\text{ min} = 45\text{ min} \times \frac{1\text{ h}}{60\text{ min}} = \frac{3}{4}\text{ h}$
(i) $W = ? \quad\quad$ (ii) cost of using $= ?$

(i) $\begin{aligned}[t] W &= \frac{V^2}{R} t \\ &= \frac{240^2}{60} \times \frac{3}{4} = 720\text{ Wh} \\ &= 720\text{ Wh} \times \frac{1\text{ kWh}}{1000\text{ Wh}} \\ &= \mathbf{0.72\text{ kWh}} \end{aligned}$

(ii) $1\text{ kWh} = 1\text{ unit of electricity}$
$0.72\text{ kWh} = 0.72\text{ unit}$
Cost of using $0.72\text{ kWh}$
$\begin{aligned}[t] &= 0.72 \times 35 = \mathbf{25.2\text{ kyats}} \end{aligned}$

$V = 220\text{ V},\quad I = 10\text{ A},\quad t = 6\text{ h}$
(i) $W = ? \quad\quad$ (ii) cost of using $= ?$

(i) $\begin{aligned}[t] W &= V I t \\ &= 220 \times 10 \times 6 = 13200\text{ Wh} \\ &= 13200\text{ Wh} \times \frac{1\text{ kWh}}{1000\text{ Wh}} \\ &= \mathbf{13.2\text{ kWh}} \end{aligned}$

(ii) $1\text{ kWh} = 1\text{ unit of electricity}$
$13.2\text{ kWh} = 13.2\text{ unit}$
Cost of using $13.2\text{ kWh}$
$\begin{aligned}[t] &= 50 \times 13.2 = \mathbf{660\text{ kyats}} \end{aligned}$

$t = 50\text{ min} = 50 \times \frac{1\text{ h}}{60\text{ min}} = \frac{5}{6}\text{ h}$, $W = ?$, $\text{payment} = ?$
$P = \frac{W}{t}$
$W = P t = 1200 \times \frac{5}{6} = 1000\text{ Wh} \times \frac{1\text{ kWh}}{1000\text{ Wh}} = 1\text{ kWh}$
$1\text{ kWh} = 1\text{ unit of electricity}$

The electricity meter reading increases
$\mathbf{1\text{ unit of electricity}}$.
$\text{The payment} = 1 \times 35 = \mathbf{35\text{ kyats}}$

$R_1 = 5\text{ }\Omega$, $R_2 = 10\text{ }\Omega$, $R_3 = 15\text{ }\Omega$, $P = 0.5\text{ W}$

$\begin{aligned}[t] P &= \frac{V^2}{R} \implies V^2 = P R \\ V_{\max} &= \sqrt{P R_{\min}} = \sqrt{0.5 \times 5} = \sqrt{2.5} = \mathbf{1.58\text{ V}} \end{aligned}$
For parallel combination, $V = V_1 = V_2 = V_3 = 1.58\text{ V}$
By using Ohm's law ($I = \frac{V}{R}$):
$\begin{aligned}[t] I_1 &= \frac{V_1}{R_1} = \frac{1.58}{5} = \mathbf{0.316\text{ A}} \\ I_2 &= \frac{V_2}{R_2} = \frac{1.58}{10} = \mathbf{0.158\text{ A}} \\ I_3 &= \frac{V_3}{R_3} = \frac{1.58}{15} = \mathbf{0.105\text{ A}} \end{aligned}$

$P = 60\text{ W}$, $V = 220\text{ V}$
(i) $I = ?$, (ii) $R = ?$, (iii) $Q = ?\text{ } (t = 1\text{ min} = 60\text{ s})$, (iv) $H = ?\text{ } (t = 1\text{ min} = 60\text{ s})$

$\begin{aligned}[t] \text{(i) } I &= \frac{P}{V} = \frac{60}{220} = \mathbf{0.2727\text{ A}} \\ \text{(ii) } P &= \frac{V^2}{R} \implies R = \frac{V^2}{P} = \frac{220 \times 220}{60} = \mathbf{806.67\text{ }\Omega} \\ \text{(iii) } Q &= I t = 0.2727 \times 60 = \mathbf{16.36\text{ C}} \\ \text{(iv) } H &= \frac{V I t}{J} = \frac{P t}{J} = \frac{60 \times 60}{4.2} = \mathbf{857.14\text{ cal}} \end{aligned}$

$P = 1200\text{ W}$, $V = 220\text{ V}$

$\begin{aligned}[t] \text{(i) } R &= \frac{V^2}{P} = \frac{220 \times 220}{1200} = \mathbf{40.33\text{ }\Omega} \\ \text{(ii) } I &= \frac{P}{V} = \frac{1200}{220} = \mathbf{5.45\text{ A}} \\ \text{(iii) } H &= \frac{P t}{J} = \frac{1200 \times 1}{4.2} = \mathbf{285.71\text{ cal}} \quad (t = 1\text{ s}) \\ \text{(iv) } P &= \frac{V^2}{R} = \frac{180 \times 180}{40.33} = \mathbf{803.37\text{ W}} \quad (V = 180\text{ V}) \end{aligned}$

$I = 3\text{ A}$, $V = 220\text{ V}$, $t = 1\text{ min} = 60\text{ s}$, $H = ?$

$\begin{aligned}[t] H &= \frac{V I t}{J} \\ &= \frac{220 \times 3 \times 60}{4.2} \\ &= 9428.57\text{ cal} \\ &= 9428.57 \times \frac{1\text{ kcal}}{1000\text{ cal}} = \mathbf{9.43\text{ kcal}} \end{aligned}$

$R = 2\text{ }\Omega$, $E = 6\text{ V}$, $r = 1\text{ }\Omega$, $t = 1\text{ s}$

By circuit equation,
$\begin{aligned}[t] I &= \frac{E}{R + r} = \frac{6}{2 + 1} = 2\text{ A} \\ \text{Rate of heat } \frac{H}{t} &= \frac{I^2 R}{J} = \frac{2^2 \times 2 \times 1}{4.2} = \mathbf{1.905\text{ cal s}^{-1}} \end{aligned}$

$R_1 = 2\text{ }\Omega$, $R_2 = 3\text{ }\Omega$, $R_3 = 4\text{ }\Omega$, $E = 12\text{ V}$, $r = 1\text{ }\Omega$

$R_2$ and $R_3$ are connected in parallel:
$\begin{aligned}[t] \frac{1}{R_p} &= \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} \\ R_p &= \frac{12}{7} = 1.71\text{ }\Omega \end{aligned}$
$R_p$ and $R_1$ are connected in series:
$R = R_1 + R_p = 2 + 1.71 = 3.71\text{ }\Omega$
By circuit equation:
$\begin{aligned}[t] I &= \frac{E}{R + r} = \frac{12}{3.71 + 1} = \frac{12}{4.71} = 2.55\text{ A} \end{aligned}$
By Joule's law of electricity and heat:
$\begin{aligned}[t] \frac{H}{t} &= \frac{I^2 R}{J} \text{ (by the equivalent resistors)} \\ &= \frac{2.55 \times 2.55 \times 3.71}{4.2} = \mathbf{5.744\text{ cal s}^{-1}} \end{aligned}$

$R = 10\text{ }\Omega$, $E_1 = 6\text{ V}$, $r_1 = 0.5\text{ }\Omega$, $E_2 = 6\text{ V}$, $r_2 = 0.5\text{ }\Omega$, $E_3 = 12\text{ V}$, $r_3 = 1\text{ }\Omega$, $t = 10\text{ min} = 600\text{ s}$

The batteries are connected in series aiding,
$\begin{aligned}[t] E &= E_1 + E_2 + E_3 = 6 + 6 + 12 = 24\text{ V} \\ r &= r_1 + r_2 + r_3 = 0.5 + 0.5 + 1 = 2\text{ }\Omega \end{aligned}$
By circuit equation:
$\begin{aligned}[t] I &= \frac{E}{R + r} = \frac{24}{10 + 2} = 2\text{ A} \end{aligned}$
By Joule's law of electricity and heat:
$\begin{aligned}[t] H &= \frac{I^2 R t}{J} \text{ (by } 10\text{ }\Omega \text{ resistor)} \\ &= \frac{2 \times 2 \times 10 \times 600}{4.2} = \mathbf{5714.28\text{ cal}} \end{aligned}$

$E = 12\text{ V}$, $r = 2\text{ }\Omega$, $R_1 = 1\text{ }\Omega$, $R_2 = 1\text{ }\Omega$, $R_3 = 1.5\text{ }\Omega$

$R_1$ and $R_2$ are connected in parallel:
$\begin{aligned}[t] \frac{1}{R_p} &= \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{1} + \frac{1}{1} = 2 \implies R_p = 0.5\text{ }\Omega \end{aligned}$
$R_p$ and $R_3$ are connected in series:
$R = R_p + R_3 = 0.5 + 1.5 = 2\text{ }\Omega$
By circuit equation:
$\begin{aligned}[t] I &= \frac{E}{R + r} = \frac{12}{2 + 2} = 3\text{ A} \end{aligned}$
By Joule's law of electricity and heat:
$\begin{aligned}[t] \frac{H}{t} &= \frac{I^2 r}{J} \text{ (in the battery)} \\ &= \frac{3 \times 3 \times 2}{4.2} = \mathbf{4.286\text{ cal s}^{-1}} \end{aligned}$

$P = 1200\text{ W}$, $V = 220\text{ V}$

$\begin{aligned}[t] P &= V I \implies I = \frac{P}{V} = \frac{1200}{220} = 5.45\text{ A} \end{aligned}$
$1200\text{ W}$ electric stove will draw $5.45\text{ A}$ to operate.
The rating of the fuse should be $6\text{ A}$.

$I_{\max} = 3\text{ A}$, $V = 220\text{ V}$

$\begin{aligned}[t] P_{\max} &= V I_{\max} \\ &= 220 \times 3 = \mathbf{660\text{ W}} \end{aligned}$

$I_{\max} = 5\text{ A}$, $V = 220\text{ V}$, $P_{\max} = ?$

$\begin{aligned}[t] P_{\max} &= V I_{\max} \\ &= 220 \times 5 = \mathbf{1100\text{ W}} \end{aligned}$

$I_{\max} = 5\text{ A}$, $V = 220\text{ V}$
$\begin{aligned}[t] P_{\max} &= V I_{\max} = 220 \times 5 = 1100\text{ W} \end{aligned}$

Electrical power of twenty $60\text{ W}$ electric lamps:
$P_{\text{used}} = 20 \times 60 = 1200\text{ W}$
Since $P_{\text{used}} > P_{\max}$,
No, twenty $60\text{ W}$ electric lamps cannot be used at the same time in that circuit.

$I_{\max} = 10\text{ A}$ (fuse), $V = 220\text{ V}$

$\begin{aligned}[t] P_{\max} &= V I_{\max} \\ &= 220 \times 10 = \mathbf{2200\text{ W}} \end{aligned}$
Electrical power of Ten $100\text{ W}$ electric lamps and two $150\text{ W}$ refrigerators,
$\begin{aligned}[t] P_{\text{used}} &= [10 \times 100] + [2 \times 150] = 1300\text{ W} \end{aligned}$

(i)
$\begin{aligned}[t] P &= \frac{W}{t} \\ W &= P_{\text{used}} \times t \\ &= 1300 \times 10 = 13,000\text{ Wh} \\ &= 13,000\text{ Wh} \times \frac{1\text{ kWh}}{1000\text{ Wh}} \\ &= \mathbf{13\text{ kWh}} \end{aligned}$
$1\text{ kWh} = 1\text{ unit of electricity}$
$13\text{ kWh} = 13\text{ units}$
$\begin{aligned}[t] \text{Cost of using } 13\text{ kWh} &= 50 \times 13 = \mathbf{650\text{ kyats}} \end{aligned}$

(ii)
Remaining power
$\begin{aligned}[t] &= P_{\max} - P_{\text{used}} = 2200 - 1300 = 900\text{ W} \\ n &= \frac{\text{Remaining power}}{60} = \frac{900}{60} = \mathbf{15\text{ lamps}} \end{aligned}$
The maximum number of $60\text{ W}$ electric lamps which can be safely used in addition is **$15\text{ lamps}$**.