A battery consists of 10 dry cells each of end 1.5V and internal resistance 0.5V the terminals of the battery are connected in a parallel combination of 3 omega and 5 omega find the current passing through battery
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8.5V
Given: A battery of emf 10V and internal resistance 3Ω is connected to a resistor. The current in the circuit is 0.5A
To find the terminal voltage of the battery when the circuit is closed
Solution:
Emf of the battery, E=10V
Internal resistance of the battery, r=3Ω
Current in the circuit, I=0.5A
Resistance of the resistor = R
The relation for current using Ohm's law is,
I=R+rE⟹R+r=IE⟹R+r=0.510⟹R=20−3=17Ω
Terminal voltage of the resistor = V
According to Ohm's Law,
V=IR⟹V=0.5×17⟹V=8.5V
Therefore, the resistance of the resistor is 17Ω and the terminal voltage is 8.5V
Explanation:
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