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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