The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3 Ω , 9 Ω and 9 Ω and a capacitor 5.0 μ F. How much is the current I in the circuit in steady state ?

Question

The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3 Ω , 9 Ω and 9 Ω and a capacitor 5.0 μ F. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e476fa3b531954d9dad9e56b18730f2f-.png" /> How much is the current I in the circuit in steady state ? (A) 1.6 A (B) 0.67 A (C) 2.5 A (D) 0.25 A

Tardigrade Admin · Accepted Answer

In steady state capacitor is fully charged hence no current will flow through line

2

. By simplyfing the circuit
Hence resultant potential difference across resistances will be

8.0 V .

Thus current

I = \frac{V}{R} = \frac{8.0}{3 + 9} = \frac{8}{12}

or,

I = \frac{2}{3} = 0.67 A

Q. The circuit shown here has two batteries of 8.0V and 16.0V and three resistors 3Ω,9Ω and 9Ω and a capacitor 5.0μF. How much is the current I in the circuit in steady state ?

1.6 A

0.67 A

2.5 A

0.25 A

In steady state capacitor is fully charged hence no current will flow through line 2. By simplyfing the circuit Hence resultant potential difference across resistances will be 8.0V. Thus current I=RV​=3+98.0​=128​ or, I=32​=0.67A

Q. The circuit shown here has two batteries of $8.0 V$ and $16.0 V$ and three resistors $3Ω, 9Ω$ and $9Ω$ and a capacitor $5.0 μ F$ .

How much is the current $I$ in the circuit in steady state ?

In steady state capacitor is fully charged hence no current will flow through line $2$ . By simplyfing the circuit
Hence resultant potential difference across resistances will be $8.0 V .$
Thus current $I = \frac{V}{R} = \frac{8.0}{3 + 9} = \frac{8}{12}$
or, $I = \frac{2}{3} = 0.67 A$