A current of 1 A flows in a series circuit c | vedclass

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Question

(A) Step $1$: Calculate the resistance of the lamp. In a series circuit,total resistance $R_{total} = V / I = 10 \ V / 1 \ A = 10 \ \Omega$. Since $R_

Vedclass · Accepted Answer

Resistance of lamp is $5 \ \Omega$; no change in current or potential difference.

Vedclass · Answer

(A) Step $1$: Calculate the resistance of the lamp. In a series circuit,total resistance $R_{total} = V / I = 10 \ V / 1 \ A = 10 \ \Omega$. Since $R_{total} = R_{lamp} + R_{conductor}$,we have $10 \ \Omega = R_{lamp} + 5 \ \Omega$. Therefore,$R_{lamp} = 5 \ \Omega$. Step $2$: Analyze the parallel connection. When a $10 \ \Omega$ resistor is connected in parallel to the series combination,the potential difference across the series combination remains $10 \ V$ because the battery is connected directly across the parallel branches. Since the potential difference across the series branch (lamp + $5 \ \Omega$ conductor) remains $10 \ V$,the current flowing through the series branch remains $I = V / R_{total} = 10 \ V / 10 \ \Omega = 1 \ A$. Consequently,there will be no change in the current flowing through the $5 \ \Omega$ conductor,and there will be no change in the potential difference across the lamp.

Sample	Resistivity $(\Omega \cdot m)$
$A$	$1.6 \times 10^{-8}$
$B$	$7.5 \times 10^{17}$
$C$	$44 \times 10^{-6}$

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