A PMMC instrument has full-scale deflection current of \(50\ \mu\text{A}\) and coil resistance of \(1\ \text{k}\Omega\). The resistance to be added in series to convert it into 10 V voltmeter is
Show Hint
Alternatively, use total resistance directly: \(R_{\text{total}} = \frac{V_{\text{new}}}{I_{fs}} = \frac{10\text{ V}}{50\ \mu\text{A}} = 200\ \text{k}\Omega\). Since the meter already has an internal resistance of \(1\ \text{k}\Omega\), the added series resistance is simply \(200\ \text{k}\Omega - 1\ \text{k}\Omega = 199\ \text{k}\Omega\).