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(B) Given: Initial potential difference $V_1 = 60\, V$,Initial current $I_1 = 4\, A$.According to Ohm's law,$V = IR$,where $R$ is the resistance of th

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$8$

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(B) Given: Initial potential difference $V_1 = 60\, V$,Initial current $I_1 = 4\, A$.According to Ohm's law,$V = IR$,where $R$ is the resistance of the heater. Since the heater remains the same,its resistance $R$ is constant.First,calculate the resistance: $R = V_1 / I_1 = 60\, V / 4\, A = 15\, \Omega$.Now,for the new potential difference $V_2 = 120\, V$,the new current $I_2$ is calculated as:$I_2 = V_2 / R = 120\, V / 15\, \Omega = 8\, A$.Alternatively,since $I \propto V$,doubling the potential difference from $60\, V$ to $120\, V$ will double the current from $4\, A$ to $8\, A$.

The potential difference between the terminals of an electric heater is $60\, V$ when it draws a current of $4\, A$ from the source. What current will the heater draw,if the potential difference is increased to $120\, V$ (in $, A$)?

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