$\varepsilon I$

Potential Drop along a circuit

• $V_A=V_B+I R$

3. $I^2 R=\frac{V^2}{R}$

V= Potential difference across in the ends of resistor(Ohmic).

P = Power to be delivered.

$R_c=$ Resistance of cables.

$P_c=I^2 R_c=\frac{P^2}{{v}^2} \cdot R_c \quad$

$P=I \cdot V$

The section AC absorbs 64 watts of power.

$V_{A C}=\frac{64}{I}=\frac{64}{1.6}=40 \mathrm{~V}$

$V_{A B}=I R=1.6 \times 20=32 \mathrm{~V}$

Voltage provide by battery $=40-32=8 \mathrm{~V}$ Resistance absorbs

$I^2 R=(1.6)^2 \times 20=2.56 \times 20=51.2 \text { watts. }$

Power absorbed by the battery

64 - 51.2 = 12.8 $\mathrm{~W}$

100 Watts. -switched on for 30 days.

Household unit of electricity Rs 3/-

220-240 V

$\frac{V^2}{R}=100$

$\frac{57600}{100}=R \Rightarrow 576 \Omega$

I V = 100

$I =\frac{100}{V}=\frac{100}{240}=\frac{5}{12} \mathrm{~A}$

24 units per day.72 units (KWH) per month.

Rs216 /-

Two Bulbs 60 W , 90 W to be put in $240 \mathrm{~V}$ mains.

Series combination.

$P_1=\frac{V^2}{R_1} \Rightarrow 60=\frac{\left(240 R^2\right)}{R_1}$

$R_1=\frac{57600}{60}=960 \Omega$

$P_2=\frac{V^2}{R_2}=90 \Rightarrow R_2=640 \Omega$

$V=I R_1+I R_2 \Rightarrow I=\frac{V}{R_1+R_2}=\frac{240}{1600}=\frac{3}{20} 4$

$V_1=\frac{3}{20} \times R_1=\frac{3}{20} \times 960=144^{\circ} \mathrm{V}$

$V_2=\frac{3}{20} \times R_2=96 \mathrm{~V}$

$P_1 =I^2 \cdot R_1$

$=\frac{9}{400} \times 960=21.6 \mathrm{~W}$

$P_2 =14.4 \mathrm{~W}$

Total power $=36 \mathrm{~W}$

Parallel Combination

Voltage across the resistances are this same.

Total Power $=150 \mathrm{~W}$

$I_1=\frac{240}{960}=\frac{1}{4} \mathrm{~A}$

$I_2=\frac{240}{640}=\frac{3}{8} \mathrm{~A}$

$P_1=I_1^2 R_1=\frac{1}{16} \times 960=60 \mathrm{~W}$

$P_2=I_2^2 R_2=\frac{9}{64} \times 640=90 \mathrm{~W}$

$I=I_1+I_2$

When a potential difference is applied across the combination ( $a$ to $b$ ) the same P.D. appears across each branch.

A single resistance which can replace the combination such that I remains same.

$I_1=\frac{\Delta V}{R_1} \quad I_2=\frac{\Delta V}{R_2} \text {. }$

$I =I_1+I_2$

$=\frac{\Delta V}{R_1}+\frac{\Delta V}{R_2}$

=$\Delta V\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$

if $R_{\text {eq }}=$ Equivalent resistance.

$I=\frac{\Delta V}{R_{e q}}$

$\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_{e q}}=\sum_n \frac{1}{R_n}$

Req is smaller than the smallest resistance in the circuit.

A Same current through each member.

$\Delta V =\Delta V_1+\Delta V_2$

$=I R_1+I R_2$

• = $I\ (R_{1}+R_{2})$

$R_{e q} =\sum_i R_i$

Series NOT Pananel.

$V_{a b}=I R_1$

$V_{b c}=I R_{e q}= I \cdot \frac{R_2 R_3}{R_2+R_3}$

$V_{a c} =V_{a b}+V_{b c}$

• = $I R_1+I \frac{R_2 R_3}{R_2+R_3}$

$I =\frac{V_{a c}}{R_1+\frac{R_2 R_3}{R_2+R_3}}$

=$\frac {V_{a c} (R_2 + R_3)}{R_1 R_2+R_2 R_3+R_3 R_1}$

$R_2 || R_3$

Equivalent Resistance

$R_{e q} =\frac{R_2 R_3}{R_2+R_3}$

$=\frac{2 \times 3}{2+3}=\frac{6}{3} \Omega$

$R_{1}$ in series will $R_{eq}$

New equivalent resistance

$2+\frac{6}{5}=\frac{16}{5} \Omega$

$R_1 = 2\Omega$

$R_3 = 3\Omega$

$R_{eq} = \frac{6}{5}$

$I=\frac{8}{16 / 5}=2.5 \mathrm{~A}$

$V_{a b}=8-5=3 V=V_{c d}$

Before switch is closed.

$R_{eq} = \frac {20 \times 12}{20 + 12} = 7.5{\Omega}$

I = $\frac {21}{7.5}$ = 2.8A