Two cells of emfs E₁ = l.5 V and E₂ = 2.0 V and internal resistances r₁ = 1.0 W and r₂ = 1.5W respectively are connected to an external resistor R = 2.5W as shown in figure. Find the potential difference

(a) between points A and B

(b) between points B and C

(c) across R.

OR

An inductor of inductance 10 H and a resistor of resistance 16W are connected to a 12 V dc source.

(a) Find the final steady current.

(b) How much energy is consumed to attains this steady current?

(c) What is the power dissipated in the resistor. at this current?

We choose any direction for the current I in the circuit. If the value of I turns out td be

negative, then our choice is incorrect, the direction of the current has to be reversed. Let us choose the direction of I as shown in figure.

Applying loop rule.to loop ABCDEA,

$-E_1+I{r}_1+E_2+I{r}_2+IR=0$

$\Rightarrow \mathrm{ }-1.5+I\times 1.0+2.0+I\times 1.5+2.5I=0$

$\Rightarrow \mathrm{ }I=-0.1\mathrm{A}$

Since I is negative, our choice for the direction of I is incorrect. So the direction of I will have to be reversed as shown in figure where now I = + 0.1 A.

(a) V_A - V_B = -E₁ - I_r =- 1.5 - 0.1 x 1.0 = - 1.6 V.

The negative sign indicates that A is at a lower potential than B.

(b) V_B – V_C = E₂ - Ir₂ =2.0 - 0. l x 1.5 = l.85V.

Point B is at a higher potential that point C.

(c) V_R = IR = 0.1 x 2.5 = 0.25 V.

OR

(a)  $i_0=\frac{E}{R}=\frac{12}{16}=\frac{3}{4}\mathrm{A}=0.75\mathrm{A}$

(b) $U=\frac{1}{2}L{i}_0^2=\frac{1}{2}\times 10\times {\left(\frac{3}{4}\right)}^2=2.8\mathrm{J}$

(c)  $\mathrm{P}={\mathrm{i}}_0\mathrm{E}=\frac{3}{4}\times 12=9\mathrm{W}$

or $\mathrm{P}={\mathrm{i}}_0^2\mathrm{R}={\left(\frac{3}{4}\right)}^2\times 16=9\mathrm{W}$