In the circuit shown in the figure the batteries have neglig

In the circuit shown in the figure the batteries have negligible internal resistance and the meters are both idealized. With the switch open, the voltmeter reads 10.0V.

a)Find the EMF of the battery?

b)What does the ammeter read with the switch is closed?

c)After the switch is closed, What are the voltage drops across each resister and which direction is the current traveling through each R?

Solution

Current through 50 ohms:

I₅₀ = V/R = 10/50 = 0.2 A
Voltage across 30 ohms:

V₃₀ = I₅₀*R = 0.2 * 30 = 6 V
Current through 75 ohms:

I₇₅ = (10+6) / 75 = 0.213 A

voltage drop across 20 ohms:

V₂₀ = (0.2 + 0.213) * 20 = 8.26 V

therefore, the emf of battery:

e = 6+10 + 8.26 = 24.26 V

If S is closed, the ammeter will read:

I = 25/50 = 0.5 A