Question

find the potential difference across each cell and the rate of of energy dissipated ​

Answer 1

Given that,

Emf $\epsilon_{1}= 12\ V$

Emf $\epsilon_{2}=6\ V$

Internal resistance r₁ = 2 ohm

Internal resistance r₂ = 1 ohm

Resistance = 4 ohm

According to figure,

Applying Kirchhoff's voltage law at loop PQRSTUP

$\epsilon_{1}-\epsilon_{2}+I_{2}r_{2}-I_{1}r_{1}=0$

Put the value into the formula

$12-6+I_{2}-I_{1}2=0$

$I_{2}-I_{1}2=-6$.....(I)

Applying Kirchhoff's voltage law at loop QRSTQ

$\epsilon_{1}-(I_{1}+I_{2})R-I_{1}r_{1}=0$

Put the value into the formula

$12-(I_{1}+I_{2})4-I_{1}2=0$

$6I_{1}+4I_{2}=12$......(II)

From equation (I) and equation (II)

$I_{2}=-\dfrac{6}{7}$

Put the value into the (I)

$I_{1}=\dfrac{18}{7}$

We need to calculate the potential difference across first cell

Using formula of potential difference

$V_{1}=\epsilon_{1}-I_{1}r_{1}$

Put the value into the formula

$V_{1}=12-\dfrac{18}{7}\times2$

$V_{1}=6.85\ V$

We need to calculate the potential difference across second cell

Using formula of potential difference

$V_{2}=\epsilon_{2}-I_{2}r_{2}$

Put the value into the formula

$V_{1}=6+\dfrac{6}{7}\times1$

$V_{1}=6.85\ V$

We need to calculate the rate of of energy dissipated in R

Using formula of power consumed

$P=(I_{1}+I_{2})^2\times R$

Put the value into the formula

$P=(\dfrac{18}{7}-\dfrac{6}{7})^2\times4$

$P=11.76\ watt$

Hence, The rate of of energy dissipated in R is 11.76 watt.