Consider the circuit shown in the image. the battery has an emf 60.0 V and negli

Question

Consider the circuit shown in the image. the battery has an emf 60.0 V and negligible internal resistence. R_2 = 2 ohms, C_1= 3.00 uF, and C_2= 6.00 uF. After the capacitors have attained their final charges, the charge on C_1 is Q_! = 18.0 uC.
a)what is the final charge on C_2?
b)what is the resistance R_1?

Consider the circuit shown in the image. the battery has an emf 60.0 V and negligible internal resistence. R_2 = 2 ohms, C_1= 3.00 uF, and C_2= 6.00 uF. After the capacitors have attained their final charges, the charge on C_1 is Q_! = 18.0 uC. a)what is the final charge on C_2? b)what is the resistance R_1?

Explanation / Answer

Voltage across C1:

Q1=C1*V

V=18*10^-6/3*10^-6

=6 V

(a)

Voltage will remain same across c2:

Q2=C2*V=6*10^-6*6 = 36 C

(b)

Voltage across R2 is 6 V.

I=6/2 = 3 A

Now, by kirchoff law:

-3*R1-2*3+60=0

3R1=54

R1=18

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