Question:
A galvanometer of resistance \( G \) is converted into a voltmeter to measure up to \( V \) volts by connecting a resistance \( R_1 \) in series with the coil. If \( R_1 \) is replaced by \( R_2 \), then it can only measure up to \( \frac{V}{2} \) volts. Find the value of the resistance \( R_3 \) (in terms of \( R_1 \) and \( R_2 \)) needed to convert it into a voltmeter that can read up to 2 V.
A galvanometer of resistance \( G \) is converted into a voltmeter to measure up to \( V \) volts by connecting a resistance \( R_1 \) in series with the coil. If \( R_1 \) is replaced by \( R_2 \), then it can only measure up to \( \frac{V}{2} \) volts. Find the value of the resistance \( R_3 \) (in terms of \( R_1 \) and \( R_2 \)) needed to convert it into a voltmeter that can read up to 2 V.
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To convert a galvanometer into a voltmeter for different ranges, always relate voltage to the series resistance using \( V = I_g (R + G) \). When given multiple configurations, equate currents to find relationships between resistances.
Updated On: Jan 13, 2026
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Solution and Explanation
Step 1: General formula for converting galvanometer to voltmeter
To convert a galvanometer to a voltmeter, a resistance \( R \) is connected in series such that: \[ V = I_g (R + G) \] where \( V \) = full-scale deflection voltage, \( I_g \) = current for full-scale deflection of the galvanometer, \( G \) = resistance of the galvanometer, \( R \) = series resistance (to be calculated). Case 1: Using resistance \( R_1 \) to measure up to \( V \) \[ V = I_g (R_1 + G) \quad \cdots (1) \] Case 2: Using resistance \( R_2 \) to measure up to \( \frac{V}{2} \) \[ \frac{V}{2} = I_g (R_2 + G) \quad \cdots (2) \] Divide (1) by (2): \[ \frac{V}{\frac{V}{2}} = \frac{R_1 + G}{R_2 + G} \Rightarrow 2 = \frac{R_1 + G}{R_2 + G} \Rightarrow R_1 + G = 2(R_2 + G) \Rightarrow R_1 + G = 2R_2 + 2G \Rightarrow R_1 - 2R_2 = G \quad \cdots (3) \] Step 2: Find \( R_3 \) for 2 V full-scale deflection
Let \( R_3 \) be the series resistance needed to measure 2 V, then: \[ 2 = I_g (R_3 + G) \quad \cdots (4) \] Substitute \( I_g \) from equation (1): \[ I_g = \frac{V}{R_1 + G} \Rightarrow 2 = \frac{V}{R_1 + G} (R_3 + G) \Rightarrow R_3 + G = \frac{2(R_1 + G)}{V} \] Use equation (3): \( G = R_1 - 2R_2 \) \[ R_3 + (R_1 - 2R_2) = \frac{2(R_1 + R_1 - 2R_2)}{V} \Rightarrow R_3 = \frac{2(2R_1 - 2R_2)}{V} - (R_1 - 2R_2) \] Simplify: \[ R_3 = \frac{4(R_1 - R_2)}{V} - (R_1 - 2R_2) \] % Final Answer Statement Answer: The required resistance is \[ \boxed{R_3 = \frac{4(R_1 - R_2)}{V} - (R_1 - 2R_2)} \]
To convert a galvanometer to a voltmeter, a resistance \( R \) is connected in series such that: \[ V = I_g (R + G) \] where \( V \) = full-scale deflection voltage, \( I_g \) = current for full-scale deflection of the galvanometer, \( G \) = resistance of the galvanometer, \( R \) = series resistance (to be calculated). Case 1: Using resistance \( R_1 \) to measure up to \( V \) \[ V = I_g (R_1 + G) \quad \cdots (1) \] Case 2: Using resistance \( R_2 \) to measure up to \( \frac{V}{2} \) \[ \frac{V}{2} = I_g (R_2 + G) \quad \cdots (2) \] Divide (1) by (2): \[ \frac{V}{\frac{V}{2}} = \frac{R_1 + G}{R_2 + G} \Rightarrow 2 = \frac{R_1 + G}{R_2 + G} \Rightarrow R_1 + G = 2(R_2 + G) \Rightarrow R_1 + G = 2R_2 + 2G \Rightarrow R_1 - 2R_2 = G \quad \cdots (3) \] Step 2: Find \( R_3 \) for 2 V full-scale deflection
Let \( R_3 \) be the series resistance needed to measure 2 V, then: \[ 2 = I_g (R_3 + G) \quad \cdots (4) \] Substitute \( I_g \) from equation (1): \[ I_g = \frac{V}{R_1 + G} \Rightarrow 2 = \frac{V}{R_1 + G} (R_3 + G) \Rightarrow R_3 + G = \frac{2(R_1 + G)}{V} \] Use equation (3): \( G = R_1 - 2R_2 \) \[ R_3 + (R_1 - 2R_2) = \frac{2(R_1 + R_1 - 2R_2)}{V} \Rightarrow R_3 = \frac{2(2R_1 - 2R_2)}{V} - (R_1 - 2R_2) \] Simplify: \[ R_3 = \frac{4(R_1 - R_2)}{V} - (R_1 - 2R_2) \] % Final Answer Statement Answer: The required resistance is \[ \boxed{R_3 = \frac{4(R_1 - R_2)}{V} - (R_1 - 2R_2)} \]
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