A dry cell of emf 1.6v and internal resistance 0.1ohm is connected to a resistor of resistance R ohm.if the current drawn from the cell is 2ampere a) what is the voltage drop across R ? b) what is the energy dissipation in resistor?
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E= I(R+r)
E is EMF ,I current ,R External Resistance,r Internal Resistance
1.6=2(R+0.1)
R+0.1=0.8
R=0.7 ohm
Voltage drop across R = IR
= 2×0.7= 1.4V
------------------–------------------------------------
Heat dissipation= I²RT
= 4×0.7×T=2.8T joules
------------------–------------------------------------
Power dissipation = Voltage drop×current
= 1.4×2= 2.8 watt
hope this helps
E is EMF ,I current ,R External Resistance,r Internal Resistance
1.6=2(R+0.1)
R+0.1=0.8
R=0.7 ohm
Voltage drop across R = IR
= 2×0.7= 1.4V
------------------–------------------------------------
Heat dissipation= I²RT
= 4×0.7×T=2.8T joules
------------------–------------------------------------
Power dissipation = Voltage drop×current
= 1.4×2= 2.8 watt
hope this helps
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