Question

Please, please answer. Class 12th. Find the EMF E1 and E2 in the given network. Also find the potential difference between points A and B.

Instead of solving you can tell me how to solve it.​

Answer 1

Given that,

Resistance $R_{1}$ = 6 Ω

Resistance $R_{2}$ = 6 Ω

Resistance $R_{3}$ = 4 Ω

Resistance $R_{4}$ = 1 Ω

Resistance $R_{5}$ = 1 Ω

Resistance $R_{6}$ = 2 Ω

Voltage = 20 V

Current in abcd $I_{1}= 1\ A$

Current in abef $I_{2}= 2\ A$

We need to calculate the value of emf E₁

Using kirchhoff's law in loop abcd

$-R_{1}I_{1}-R_{2}I__{1}+V+R_{3}I_{1}-E_{1}+R_{4}I_{1}=0$

Put the value into the formula

$-6\times1-6\times1+20+4\times1-E_{1}+1\times1=0$

$-6-6+20+4-E_{1}+1=0$

$-E_{1}=-1-8-4$

$E_{1}=13\ V$

We need to calculate the value of emf E₂

Using kirchhoff's law in loop abef

$-R_{3}I_{1}+E_{1}-R_{4}I_{1}-E_{2}-R_{5}I_{2}-R_{6}I_{2}=0$

Put the value into the formula

$-4\times1+E_{1}-1\times1-E_{2}-1\times2-2\times2=0$

$E_{1}-E_{2}=4+1+2+4$

$-E_{2}=11-13$

$E_{2}=2\ V$

We need to calculate the potential difference between a and b

Using  kirchhoff's law in loop ab

$V_{B}-R_{4}I_{1}+E_{1}-R_{4}I_{1}=V_{A}$

$V_{A}-V_{B}=-4\times1+13-1\times1$

$V_{A}-V_{B}=8\ V$

Hence, The value of emf E₁ and E₂ is 13 V and 2 V.

The potential difference between A and B is 8 V.