The circuit shown in figure below consists of external resistance of 10 ohm connected across two batteries of emfs 6 V and 9 V with internal resistance 1 ohm each. Find the power dissipated by the 10 ohm resistor.
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Answered by
0
Answer:
10^10^100 is the correct answer
Explanation:
10×10=100 like this ha ha ha
Answered by
3
Answer:
Given:-
External resistance (R1) = 10ohm
Internal resistance (R2) = 1ohm
Voltage (V1) = 6V
Voltage (V2) = 9V
Vnet = V2 - V1
= 9 - 6
= 3V.
Rnet = R1+R2
= 10+1
= 11ohm.
By applying, I = V/R
= 3/11
= 0.272727Amp.
By applying, P = V^2/R
= 9/11
= 0.8181812W.
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