1. ⇒ (MHT CET 2023 13th May Morning Shift )

Two cells $E_{1}$ and $E_{2}$ having equal EMF ' $E$ ' and internal resistances $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively are connected in series. This combination is connected to an external resistance ' $R$ '. It is observed that the potential difference across the cell $E_{1}$ becomes zero. The value of ' $R$ ' will be

A. $r_{1} - r_{2}$

B. $r_{1} + r_{2}$

C. $\frac{r_{1} - r_{2}}{2}$

D. $\frac{r_{1} + r_{2}}{2}$

Correct Option is (A)

The total current in the circuit is

$I = \frac{2 E}{r_{1} + r_{2} + R}$ ... (1) .... (Given cells are in series, $E + E = 2 E)$

Now the potential drop across the first cell is $V_{1} = E - r_{1} = 0$

$\begin{aligned} ∴ & E - (\frac{2 E}{r_{1} + r_{2} + R}) \times r_{1} = 0 \\ \frac{2 E}{r_{1} + r_{2} + R} = \frac{E}{r_{1}} \Rightarrow 2 r_{1} = r_{1} + r_{2} + R \\ ∴ & R = r_{1} - r_{2} \end{aligned}$

2. ⇒ (MHT CET 2023 12th May Evening Shift )

A potentiometer wire of length $4 m$ and resistance $5 Ω$ is connected in series with a resistance of $992 Ω$ and a cell of e.m.f. $4 V$ with internal resistance $3 Ω$ . The length of $0.75 m$ on potentiometer wire balances the e.m.f. of

A. 4.00 mV

B. 3.75 mV

C. 3.00 mV

D. 2.50 mV

Correct Option is (B)

$∴$ Total Resistance:

$R = 992 + 5 + 3 = 1000 Ω$

Voltage across $4 m$ wire:

$\frac{5}{995 + 5} \times 4 = 0.02 V$

$∴$ For one metre wire:

$\frac{0.02}{4} = 0.005 V$

$∴$ For $0.75 m$ wire:

$\begin{aligned} 0.004 \times 0.75 & = 0.00375 \\ = 3.75 mV \end{aligned}$

3. ⇒ (MHT CET 2023 12th May Morning Shift )

Resistance of a potentiometer wire is $2 Ω / m$ . A cell of e.m.f. $1.5 V$ balances at $300 cm$ . The current through the wire is

A. $2.5 mA$

B. $7.5 mA$

C. $250 mA$

D. $750 mA$

Correct Option is (C)

$l = 300 cm = 3 m$

Total resistance of wire,

$R = 3 \times 2 = 6 Ω$

Since, the potentiometer is balanced. Voltage across wire segment $= 1.5 V$

$\begin{array}{ll} ∴ & IR = 1.5 V \\ ∴ & I = \frac{1.5}{6} = 250 mA \end{array}$

4. ⇒ (MHT CET 2023 12th May Morning Shift )

A potentiometer wire has length of $5 m$ and resistance of $16 Ω$ . The driving cell has an e.m.f. of $5 V$ and an internal resistance of $4 Ω$ . When the two cells of e.m.f.s $1.3 V$ and $1.1 V$ are connected so as to assist each other and then oppose each other, the balancing lengths are respectively

A. $3 m, 0.25 m$

B. $0.25 m, 3 m$

C. $2.5 m, 0.3 m$

D. $0.3 m, 2.5 m$

Correct Option is (A)

$\begin{aligned} K = \frac{E R}{(R + r) L} \\ E = 5 V, r = 4 Ω, L = 5 m, R = 16 Ω \\ ∴ K = \frac{5 \times 16}{(16 + 4) \times 5} \\ ∴ K = 0.8 V / m \end{aligned}$

When ' $E_{1}$ ' and ' $E_{2}$ ' are connected so as to assist each other

$\begin{array}{ll} E_{1} + E_{2} = K_{1} \\ 1.3 + 1.1 = 0.8 \times l_{1} \\ ∴ & l_{1} = 3 m \end{array}$

When ' $E_{1}$ ' and ' $E_{2}$ ' are connected so as to oppose each other,

$\begin{aligned} E_{1} - E_{2} = K l_{2} \\ 1.3 - 1.1 = 0.8 \times l_{2} \\ ∴ & l_{2} = 0.25 m \end{aligned}$

As, value for balancing lengths are different in all the options. It is sufficient to calculate balancing length in any one case (Assisting/ opposing) to reach the final correct answer.

5. ⇒ (MHT CET 2023 11th May Evening Shift )

Two batteries, one of e.m.f. $12 V$ and internal resistance $2 Ω$ and other of e.m.f. $6 V$ and internal resistance $1 Ω$ , are connected as shown in the figure. What will be the reading of the voltmeter 'V'?

MHT CET 2023 11th May Evening Shift Physics - Current Electricity Question 7 English

A. 12 V

B. 8 V

C. 6 V

D. 4 V

Correct Option is (B)

The formula for the equivalent emf of the parallel combination of batteries is

$ε_{e} = r_{eq} (\frac{e_{1}}{r_{1}} + \frac{e_{2}}{r_{2}})$

Here, $r_{eq}$ is the equivalent resistance

$\begin{aligned} \frac{1}{r_{eq}} = \frac{1}{r_{1}} + \frac{1}{r_{2}} \\ \frac{1}{r_{eq}} = \frac{1}{2} + \frac{1}{1} \\ \frac{1}{r_{eq}} = \frac{3}{2} \end{aligned}$

Substituting the values

$\begin{aligned} ε_{e} & = r_{eq} (\frac{e_{1}}{r_{1}} + \frac{e_{2}}{r_{2}}) \\ ε_{c} & = \frac{2}{3} (\frac{12}{2} + \frac{6}{1}) \\ ε_{e} & = \frac{2}{3} \times 12 \\ ∴ ε_{e} & = 8 V \end{aligned}$

⇒ Current Electricity