A galvanometer of resistance 100 Omega when c | vedclass

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(C) First,calculate the full-scale deflection current $i_g$ of the galvanometer.Given that the galvanometer (resistance $R_g = 100 \ \Omega$) is in se

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$20$

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(C) First,calculate the full-scale deflection current $i_g$ of the galvanometer.Given that the galvanometer (resistance $R_g = 100 \ \Omega$) is in series with $400 \ \Omega$ to measure $10 \ V$:$i_g = \frac{V}{R_g + R_{series}} = \frac{10}{100 + 400} = \frac{10}{500} = 20 	imes 10^{-3} \ A$.To convert the galvanometer into an ammeter to read up to $I = 10 \ A$,a shunt resistance $S$ is connected in parallel.The condition for the shunt is $i_g R_g = (I - i_g) S$.Substituting the values:$20 	imes 10^{-3} 	imes 100 = (10 - 20 	imes 10^{-3}) S$.Since $20 	imes 10^{-3} = 0.02 \ A$,we have:$2 = (10 - 0.02) S = 9.98 S$.$S = \frac{2}{9.98} \approx 0.2004 \ \Omega$.Given the form $x 	imes 10^{-2} \ \Omega$,we approximate $S \approx 20 	imes 10^{-2} \ \Omega$.Thus,$x = 20$.

$A$ galvanometer of resistance $100 \Omega$ when connected in series with $400 \Omega$ measures a voltage of up to $10 \ V$. The value of resistance required to convert the galvanometer into an ammeter to read up to $10 \ A$ is $x \times 10^{-2} \Omega$. The value of $x$ is:

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