Question

A galvanometer of resistance \(200 \Omega\) gives full scale deflection with 15 milli-ampere current. In order to convert it into a \(15 \mathrm{~V}\) range voltmeter. What is the value of resistance connected in series ? (A) \(1000 \Omega\) (B) \(800 \Omega\) (C) \(2500 \Omega\) (D) \(1500 \Omega\)

Short answer

The value of the series resistance needed to convert the galvanometer into a 15 V range voltmeter is \(800 \Omega\). The correct answer is (B) \(800 \Omega\).

Step-by-step solution

Recall Ohm's Law

Ohm's Law states that voltage (V) is equal to the product of current (I) and resistance (R): V = IR. We will use this equation to determine the value of the series resistance needed to convert the galvanometer into a voltmeter.

Calculate the voltage across the galvanometer

We are given that the full-scale deflection current of the galvanometer is 15 mA (or 0.015 A). To find the voltage across the galvanometer, we'll use Ohm's Law (V = IR) with the given resistance of 200 Ω: V_galvanometer = I_galvanometer × R_galvanometer V_galvanometer = (0.015 A) × (200 Ω) V_galvanometer = 3 V

Determine the voltage across the series resistance

Since we want to convert the galvanometer into a 15 V range voltmeter, the total voltage across both the galvanometer and the series resistance (R_series) must be 15 V. To find the voltage across the series resistance, we'll subtract the voltage across the galvanometer from the total voltage: V_series_resistance = V_total - V_galvanometer V_series_resistance = 15 V - 3 V V_series_resistance = 12 V

Calculate the value of the series resistance

Now we have the voltage across the series resistance and we know that the current through it is the same as the current through the galvanometer (15 mA or 0.015 A). Using Ohm's Law (V = IR), we can solve for the value of the series resistance: R_series = V_series_resistance ÷ I_galvanometer R_series = 12 V ÷ 0.015 A R_series = 800 Ω The value of the series resistance needed to convert the galvanometer into a 15 V range voltmeter is 800 Ω. The correct answer is (B) 800 Ω.