1. ⇒ (MHT CET 2023 10th May Morning Shift )

A voltmeter of resistance $150 Ω$ connected across a cell of e.m.f. $3 V$ reads $2.5 V$ . What is the internal resistance of the cell?

A. $10 Ω$

B. $15 Ω$

C. $20 Ω$

D. $30 Ω$

Correct Option is (D)

Internal resistance,

$\begin{aligned} r & = [\frac{E}{V} - 1] R \\ = [\frac{3}{2.5} - 1] 150 \\ = 30 Ω \end{aligned}$

2. ⇒ (MHT CET 2023 9th May Evening Shift )

Five current carrying conductors meet at a point ' $O$ ' as shown in figure. The magnitude and direction of the current in conductor ' $O P$ ' is

MHT CET 2023 9th May Evening Shift Physics - Current Electricity Question 18 English

A. $6.5 A$ from $O$ to $P$ .

B. $9 A$ from $P$ to $O$ .

C. $10.5 A$ from $P$ to $O$ .

D. $11.5 A$ from $O$ to $P$ .

Correct Option is (D)

Using Kirchhoff's current Law,

Current flowing in = Current flowing out $10 + 2.5 + 5 = 6 + x$

$∴ x = 17.5 - 6$

$\begin{aligned} ∴ x & = 17.5 - 6 \\ = 11.5 A \end{aligned}$

$∴$ Magnitude and direction of the current in 'OP' should be $11.5 A$ from $O$ to $P$ .

3. ⇒ (MHT CET 2021 20th September Morning Shift )

Consider the circuit shown in the figure. The value of current 'I' is

MHT CET 2021 20th September Morning Shift Physics - Current Electricity Question 25 English

A. $- \frac{7}{18}$ A

B. 5A

C. 3A

D. $- 3$ A

Correct Option is (A)

MHT CET 2021 20th September Morning Shift Physics - Current Electricity Question 25 English Explanation

Applying Kirchoff's voltage law to loop ABEFA we get

$\begin{aligned} - 28 I_{1} - 6 - 8 = 0 \\ - 28 I_{1} - 14 = 0 \\ ∴ & - 28 I_{1} = 14 \\ ∴ & I_{1} = \frac{- 14}{28} = - \frac{1}{2} A \end{aligned}$

Applying KVL to BCDEB

$\begin{aligned} 54 I_{2} - 12 + 6 = 0 \\ ∴ & 54 I_{2} = 6 \\ ∴ & I_{2} = \frac{6}{54} = \frac{1}{9} A ∴ I = I_{1} + I_{2} = - \frac{1}{2} + \frac{1}{9} = - \frac{7}{18} A \end{aligned}$

⇒ Current Electricity