The circuit in the figure below has been connected for several seconds. (Let V_1

Question

The circuit in the figure below has been connected for several seconds. (Let V_1 = 4.10 V and V_2 = 7.30 V.) (a) Find the current in the V_1 battery. 1.104 magnitude Your response differs from the correct answer by more than 10%. Double check your calculations. A direction from d toward e (b) Find the current in the 3.00-Ohm resistor. 473 magnitude Your response differs from the correct answer by more than 10%. Double check your calculations. A direction from f toward c (c) Find the current in the V_2 battery. magnitude A direction (d) Find the current in the 3.00-V battery. magnitude A direction (e) Find the charge on the capacitor. mu C

Explanation / Answer

V1 = 4.1 V and V2 = 7.3 V

from junction rule at c

I3 = I1 + I2 (1)

from loop rule in bcfgb

7.3 = 5I1 - 3I2 (2)

In loop cdefc

4.1 = 5I3 + 3I2

4.1 = 5(I1 + I2) + 3 I2 [using 1 here]

4.1 = 5 I1 + 8 I2 (3)

(3) - (2) gives us

- 11 I2 = 3.2 => I2 = -0.291 A

I1 = (4.1 - 8I2)/5 = 1.286 A

I3 = 1.286 - 0.291 = 0.995 A

a)The current in V1 is = 0.995 A, down

b)Current in 3 Ohm, I2 = -0.291 A down

c)Current in V2 is I1 = 1.286 A, up

d)current in 3V is 0 since voltage drop due to capacitor.

e)Q = CV = 6 x 10^-6 x (3 + 7.3) = 61.8 uF

Hence, Q = 61.8 uF

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