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# What is the steady state potential (in V) of point a with respect to point b in figure when switch S is open?

Last updated date: 15th Aug 2024
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Hint: In a steady state, the capacitor becomes open. So, no current will flow through them. We also know, no current flows through an open switch. Thus, no current will flow through capacitors as well as open switches. Using this information, find the potential at point a and then the potential at point b. Subtract these potentials to get the steady state potential of point a with respect to point b in figure when switch S is open.

As it is a steady state, the capacitors are fully charged and they behave as an open circuit. Current does not pass through an open circuit. Thus, no current will pass through the capacitors. It is given that the switch S is open, so the current will only pass through $6\Omega$ resistor up to point a and will not flow through the remaining circuit.
Thus, the voltage at point a is given by,
${V}_{a}= 18 V$
Voltage at point b will be,
${V}_{b}=0$
Potential of point a with respect to point b is given by,
${V}_{ab}= {V}_{a}-{V}_{b}$
Substituting the values in above equation we get,
${V}_{ab}=18-0$
$\Rightarrow {V}_{ab}= 18 V$
Thus, the steady state potential (in V) of point a with respect to point b in figure when switch S is open is 18 V.

Note:
Students should not get confused between an open and a close circuit. Current passes through a closed circuit but it does not flow through an open circuit. The term steady state is a state of circuit in which the charge or current flowing into any point in the circuit is equal to the charge or current flowing out of it. A system can attain steady state when the current at each point in the circuit is constant i.e. the current does not change with time.