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Grade 12Electric Current

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

9 Years agoGrade 12

1 Answer

Prasanna Vibhandik

9 Years ago

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= $(\frac{V}{R+r})^{2} . R$

The fig. Shows a network of resistance in which the point A is earthed

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