Question

The setup in the figure is used to measure resistance R. The ammeter and voltmeter resistance are $0.01\Omega$ and $2000\Omega$ respectively. Their readings are 2 A and 180 V respectively, giving a measured resistance of $90\Omega .$ The percentage error in the measurement is _______.

• A. -4.5

Answer 1

The correct option is A -4.5
$\text{Measured value of resistance =}90\Omega$

$\text{Resistance of voltmeter,}{R}_v=2000\Omega$

$\text{Voltage across voltmeter, V = 180 V}$

$\text{Curent through voltmeter,}$

$=\frac{V}{R_V}=\frac{180}{2000}=0.09A$

$\text{Curent through resistance, R}$

$I_R=2-I_V=(2-0.09)A$

$\Rightarrow$ $I_R=1.91A$

$\text{True value of resistance,}$

$\frac{V}{I_R}=\frac{180}{1.91}=94.24\Omega$

$\%error=\frac{90-94.24}{94.24}\times 100=-4.5\%$