Problem 32 - A Complete Resource Book in Physics for JEE Main

Question

A galvanometer of resistance \(400 \Omega\) can measure a current of \(1 \mathrm{~mA}\). To convert it into a voltmeter of range \(8 \mathrm{~V}\), the required resistance is (A) \(4600 \Omega\) (B) \(5600 \Omega\) (C) \(6600 \Omega\) (D) \(7600 \Omega\)

Step-by-Step Solution

Answer

The required resistance to convert the galvanometer into a voltmeter with a range of 8 V is 7600 Ω. Therefore, the correct answer is (D) 7600 Ω.

Step 1: Understand the given information

We know that 1. The resistance of the galvanometer (G) is 400 Ω. 2. The maximum current the galvanometer can measure (I) is 1 mA. 3. The range of the voltmeter (V) is 8 V. We need to find the required resistance (R) to convert the galvanometer into a voltmeter.

Step 2: Use Ohm's Law

Ohm's Law states that \(V = IR\), where V is the voltage, I is the current, and R is the resistance. First, let's calculate the galvanometer's total resistance when it is used as a voltmeter. To do this, we can rearrange Ohm's Law to solve for the total resistance: \[R = \frac{V}{I}\] Now, plug in the given values for V (8 V) and I (1 mA or 0.001 A): \[R = \frac{8}{0.001}\]

Step 3: Calculate the total resistance

Calculate R: \[R = 8000 \Omega\] The total resistance needed for the voltmeter is 8000 Ω.

Step 4: Find the required resistance

Now, we have to find the required resistance to be added in series with the galvanometer. Since the galvanometer and the required resistance are connected in series, their resistances need to add up to the total resistance. \[R_{req} = R_{total} - R_{galvanometer}\] Plug in the values for the total resistance (8000 Ω) and the given resistance of the galvanometer (400 Ω): \[R_{req} = 8000 - 400\]

Step 5: Calculate the required resistance

Calculate \(R_{req}\): \[R_{req} = 7600 \Omega\] The required resistance to convert the galvanometer into a voltmeter with a range of 8 V is 7600 Ω. Therefore, the correct answer is (D) 7600 Ω.