Question

In an RC circuit, if we changed the resistance to one with a higher value, we would expect that:

The area under the curve change

Capacitor discharges faster

Capacitor takes longer to achieve Qmax

Changes the voltage Vc when the capacitor charges

Answer #1

When capacitor is charging, q(t)=Qmax{1-e^[-t/(RC)]}, where q(t) is charge at time t, R is resistance and C is capacitance.

So, from the above equation , we know that if resistance R is increased, we will require longer time to reach Qmax. So, third option is correct.

Note that area under the curve represents total charge transferred which is same even if resistance value is changed. So , first option is not correct.

When discharging a capacitor,q(t)=Qmax*e^[-t/(RC)], where q(t) is charge at time t, R is resistance and C is capacitance. So, if resistance is increased, it takes more time to discharge the capacitor. So, second option is incorrect.

Also, Vc depends only on the external voltage supplied and not on resistance. So, fourth option is incorrect.

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capacitors in your RC circuit until the potential difference on the
capacitor saturates at some initial value supplied by the power
supply. Then you start to discharge your capacitor and read the
potential difference on the capacitor in at different times. When
you plot your data as lnV versus time, you make a linear fit which
looks like the one on page 5 of your lab sheet.
If the slope...

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capacitors in your RC circuit until the potential difference on the
capacitor saturates at some initial value supplied by the power
supply. Then you start to discharge your capacitor and read the
potential difference on the capacitor in at different times. When
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looks like the one on page 5 of your lab sheet.
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