JEE Main 2022 — Current Electricity Question with Solution

Shortcut Method :

Internal Resistance of Unknown Battery

$r=\left(\frac{{\ell }_1-{\ell }_2}{{\ell }_2}\right)\mathrm{R}$

Where l₁ means balanced length when key K₁ is open

Where l₂ means balanced length when key K₁ is closed

Here l₁ = 120 cm and l₂ = 80 cm and R = 4 $\Omega$

$$\therefore$$ $$r=\left(\frac{120-80}{80}\right)\times 4$$ = 2 $\Omega$

Normal Method :

Original Potentiometer is $$\to$$

Here, AB is called potentiometer wire. Let length of AB = l and resistance = R

$$\therefore$$ (1) Current in the circuit (i) = $$\frac{4}{R+R_h}$$

(2) Potential difference between A and B is

= V_A $$-$$ V_B = $$\Delta$$V_AB = i $$\times$$ R

(3) $$\frac{\Delta V_{AB}}{AB}=\frac{Potentialdifference}{Length}$$ = Potential Gradient

$$\Rightarrow \frac{\Delta V_{AB}}{l}=\frac{iR}{l}=\frac{4}{R+R_h}\times \frac{R}{l}$$ Volt/m

Now, a circuit added to potentiometer wire.

When K₁ is open then circuit look like the following.

When null point is found at C then no current will flow through the circuit added to the potentiometer wire.

Potential difference between A and C is

(1) From potentiometer wire :

$$V_A-V_C=\Delta V_{AC}=\frac{V_{AB}}{l}\times 120=\frac{4R}{(R+R_h)l}\times 120$$ ...... (1)

(2) From external circuit :

V_A $$-$$ 1.5 $$-$$ 0 $$\times$$ r = V_C

$$\Rightarrow$$ V_A $$-$$ V_C = 1.5V = $$\Delta$$V_AC ...... (2)

Comparing equation (1) and (2), we get

$$\frac{4R}{(R+R_h)l}\times 120=1.5$$ ..... (3)

When the key K₁ is closed then circuit look like this $$\to$$

When null point is found at C then no current will flow through the circuit added to the potentiometer wire but i₁ current will flow through circuit created using 1.5V with internal resistance r and 4$$\Omega$$ resistance.

Potential difference between A and C is

(1) From potentiometer wire :

$$V_A-V_C=\Delta V_{AC}=\frac{V_{AB}}{l}\times 80=\frac{4R}{(R+R_h)l}\times 80$$ ....... (4)

(2) From external circuit :

$$i_1=\frac{1.5}{4+r}$$

$$V_A-1.5+i_{1}r=V_C$$

$$\Rightarrow V_A-V_C=1.5-i_{1}r=1.5-\frac{1.5r}{4+r}$$ ..... (5)

Or we can use this also,

$$V_A-i_1\times 4=V_C$$

$$\Rightarrow V_A-V_C=i_1\times 4=\frac{1.5}{4+r}\times 4$$

From equation 4 and 5, we get

$$\frac{4R}{(R+R_h)l}\times 80=1.5-\frac{1.5r}{4+r}$$

$$\Rightarrow \frac{1.5}{120}\times 80=1.5-\frac{1.5r}{4+r}$$ [From equation 3]

$$\Rightarrow 1.5\times \frac{2}{3}=1.5-\frac{1.5r}{4+r}$$

$$\Rightarrow \frac{1.5r}{4+r}=1.5-1.5\times \frac{2}{3}$$

$$\Rightarrow \frac{1.5r}{4+r}=1.5\times \frac{1}{3}$$

$$\Rightarrow 3r=4+r$$

$$\Rightarrow 2r=4$$

$$\Rightarrow r=2\Omega$$

Other Method :

Now this can be converted to single battery and single resistance using combination of battery rule,

We know,

$$E_{eq}=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}$$

and $$r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$$

$$\therefore$$ Here, $$E_{eq}=\frac{1.5\times 4+0\times r}{4+r}=\frac{6}{4+r}$$

and $$r_{eq}=\frac{4r}{4+r}$$

$$\therefore$$ Circuit becomes

When null point is found at C then no current will flow through the circuit added to the potentiometer wire.

Potential difference between A and C is

(1) From potentiometer wire :

$$V_A-V_C=\Delta V_{AC}=\frac{V_{AB}}{l}\times 80=\frac{4R}{(R+R_h)l}\times 80$$ .... (4)

(2) From external circuit :

$$V_A-\frac{6}{4+r}-0\times \frac{4r}{4+r}=V_C$$

$$\Rightarrow V_A-V_C=\frac{6}{4+r}$$ ...... (5)

From equation 4 and 5, we get

$$\frac{4R}{(R+R_h)l}\times 80=\frac{6}{4+r}$$

$$\Rightarrow \frac{1.5}{120}\times 80=\frac{6}{4+r}$$ [From equation 3]

$$\Rightarrow 1.5\times \frac{2}{3}=\frac{6}{4+r}$$

$$\Rightarrow 4+r=6$$

$$\Rightarrow r=2\Omega$$