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**Part A**

What is the magnitude of the current in the 11 Ω resistor in (Figure 1) ?

**Part B**

What is the direction of the current in the 11 Ω resistor in (Figure 1)?

SOLUTION :Part a.Voltage across resistor of 11 Ω

= sum of emf of battery 1 and battery 2

= 3 + 6 = 9 V

So, current in the circuit = 9⁄11 = 0. 8182 amps.

Part b.Direction of current is from left to right through the resistor.

a)

Use ohm’s law

i = E1 + E2 / R

i = (6 + 3) /11 = 0.8182 A

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b)

Option (a) is correct

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If you have any doubts, please comment before the rate. I will be sure to reply. Good luck