To find the potential difference across the resistor in an RC circuit 10 seconds after the switch is closed, we need to consider the charging and discharging behavior of the capacitor.
In the given circuit, when the switch is closed, the capacitor starts to charge. The charging process follows an exponential curve described by the equation:
Vc(t) = E * (1 - e^(-t/(R*C)))
where:
Vc(t) is the voltage across the capacitor at time t,
E is the voltage of the source (30 V),
R is the resistance (1 MOhm),
C is the capacitance (5 uF),
t is the time.
After a long enough time has passed, the capacitor is fully charged, and the voltage across it reaches the source voltage E. At this point, no current flows through the resistor, and the potential difference across the resistor is zero.
To find the potential difference across the resistor 10 seconds after the switch is closed, we need to calculate Vc(10s) and subtract it from the source voltage E.
Substituting the given values into the equation:
Vc(10s) = 30 V * (1 - e^(-10s/(1 MOhm * 5 uF)))
Calculating the result:
Vc(10s) ≈ 29.534 V
Therefore, the potential difference across the resistor 10 seconds after the switch is closed is approximately E - Vc(10s) = 30 V - 29.534 V ≈ 0.466 V.
The closest answer choice from the options provided is A) 0.5 V.