Question

Find the potential difference between points A and B and between points B and C of the figure in steady state.

Answer 1

Hint: Remember the statement given in the question that circuit is in steady state which means i = 0 A find the net capacitance for AB and BC circuit and then use the formula $$(V_A-V_B)C_1=(V_B-V_C)C_2$$ to find the potential difference between A and B also B and C.

Complete Step-by-Step solution:
Let first 3uF of AB circuit be x and y and let 1uf of AC be z
Since, according to the question the circuit is in the steady state, therefore, the current in the circuit is 0 and also the circuit will be as given below

So, since the capacitor x and y are in the parallel formation .i.e. therefore the net capacitance is $$C_{net}=3+3=6uF$$
The formed circuit is

Now let 1uF of BC circuit be n and m.
Since n and m are also in the parallel formation therefore the net capacitance is $$C_{net}=1+1=2uF$$
Hence, the circuit will be

So now for the potential difference at point A and B
Since $V_A=-100$
By the formula of $$(V_A-V_B)C_1=(V_B-V_C)C_2$$
$$(-100-V_B)6=(V_B-0)2$$
$V_B=-75$
Therefore $$(V_A-V_B)=-100-(-75)$$
$$(V_A-V_B)=-25V$$
For B and C
Since we have $V_B=-75$and $V_C=0$
Therefore the potential difference between B and C is
$V_B-V_C=-75-0$
Hence $V_B-V_C=-75V$
Hence the potential difference between A and B is -25V and the potential difference between B and C is - 75V.

Note: The term steady state is a state of circuit in which the charge (or current) flowing into any point in the circuit has to equal the charge (or current) flowing out. A system can achieve a steady state when the current at each point in the circuit is constant (does not change with time).