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A dry cell of emf 1.6V and internal resistance 0.1 ohm is connected to a resistor of resistance R ohm . If the current drawn from the cell is 2A, then (i) What is the voltage drop across R? (ii) What is the energy dissipation in the resistor?

Anshuman Mishra , 8 Years ago
Grade 12
anser 1 Answers
Prasanna Vibhandik
first of all ;
The current flowing through the circuit will be:
\frac{V}{R+r}
(As the net resistance is R+r)
Now the potential drop across the resistor R ohm will be:
V – Ir
(where r is internal resistance)
SO, the energy dissipated in the ‘R’ resistor would be 
 
I2.R=  (\frac{V}{R+r})^{2} . R
 
 
 
 
 
Last Activity: 8 Years ago
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