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## Homework Statement

a. What is the potential difference V_ad in the circuit of the figure?

b. What is the terminal voltage of the 4.00 - V battery?

c. A battery with emf 10.3 V and internal resistance 0.50 ohm is inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00 - V battery. What is the difference of potential V_bc between the terminals of the 4.00 - V battery now?

## Homework Equations

See below.

## The Attempt at a Solution

I really need help with my signs. Please direct me if I am incorrect on the different parts.

a. Setting I as clockwise,

-0.50 ohm*(I) – 4.00 V – 9.00 ohm*(I) + 8.00 V – 0.5 ohm*(I) – 8.00 ohm*(I) – 6.00 ohm*(I) = 0

I = (-4.00 V)/(-24 ohm) = 0.16667 A ???

Using this,

Vad = -8 V + (8.50 ohm)*I = -8 V + (1.417 V) = -6.583 V???

b. V_bc for battery = -E + I*r = -4.00 V + (0.50 ohm)*(0.16667 A) = -3.917 V???

c. The new I must be found? Therefore,

-0.50 ohm*(I) – 4.00 V - 9.00 ohm*(I) - 0.50 ohm*(I) – 10.3 V + 8.00 V - 8.00 ohm*(I) - I(6.00 ohm) - 0.50 ohm*(I) = 0

I*(-24.5 ohm) = 6.3 V

I = 6.3 V/-24.5 ohm = -0.2571 A ??? Thus, the current actually flows counterclockwise?

V_bc = -E + I*r = -4.00 V + (0.50 ohm)*(0.2571A) = -3.8714 V???

Thank you.

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