Question

The circuit in the figure has been connected for several seconds. Find...

(a) the current in the 4V battery

(b) the current in the 3.0 resistor

(c) the current in the 8V battery

(d) the current in the 3V battery

(e) the charge in the capacitor

9.11 x 10-31 kg k=9.0 x 109 N m2/C2 7x 10-27 8.854 x 10-12 C2/Nm2 r several seconds. 几 3.0v 8 v 5 s.

Answer 1

The circuit has been connected for several seconds means that the capacitor is fully charged. The current to a fully charged capacitor is zero because the voltage V=Q/C will balance out the external voltage pushing charge onto the capacitor. After several seconds the capacitor can be replaced by a break in the circuit.

Then the circuit would be,

We have,

Applying KVL in loop 1 we get,

$5I + 3I_1= 4 \Rightarrow \boxed{8I_1 + 5 I_2 = 4}$ where, we have used the value of from previous equation.

Applying KVL in loop 2 we get,

$\boxed{-3I_1 + 5I_2 = 8}$

Solving these two boxed equation we get,

$(8I_1+5I_2) - (-3I_1 + 5I_2) = 4-8 \\ \Rightarrow 11I_1 = -4 \Rightarrow \boxed{I_1 = - \frac{4}{11} \, \text{A}}$

By putting this value in any one of the boxed equation we get,

$5I_2= 4-8I_1 = 4 + \frac{32}{11} \Rightarrow \boxed{I_2 = \frac{76}{55} \hspace{2mm} \text{A}}$

$I = I_1 + I_2 = \frac{76}{55} - \frac{4}{11} \Rightarrow \boxed{I = \frac{56}{55} \, \text{A}}$

a) The current in the 4V battery = I = 56/55 A along the direction shown in the circuit

b) The current in the 3 ohm resistor = I₁ = 4/11 A along a direction opposite to the direction shown in the circuit

c) The current in the 8V battery = I₂ = 76/55 A along the direction shown in the circuit

d) The current in the 3V battery is zero.

e) The voltage of the capacitor is 3V becasue it is directly connected to that battery, therefore the potential of the capacitor plate and the battery would be same. Hence the charge stored in the capacitor is given by,

$Q = VC = 3 \times 6 = 18\, \, \mu\text{C}$