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Question
Explain, why is the p.d. between the terminals of a storage battery less when it is supplying current than when it is on open circuit. A battery of e.m.f. 10 volts and internal resistance 2.5 ohms has two resistances of 50 ohms each connected to it. Calculate the power dissipated in each resistance
(a) When they are in series,
(b) When they are in parallel.
In each case calculate the power dissipated in the battery.
Solution
In an open circuit no current is drawn from the cell whereas in closed circuit, an amount of energy is spent in the flow of unit positive charge through the electrolyte of the cell. Thus, the p.d. between the terminals of a storage battery is less when it is supplying current.
Given, emf e = 10 V, internal resistance r = 2.5 Ω
Two external resistances R1 = R2 = 50Ω (= R, say)
(a) When R1 and R2 are connected in series, same current flows through each resisto and hence same power is dissipated in each resistor.
Total resistance of the circuit, Rs=R1 + R2 +r = 50 + 50 + 2.5 = 102.5Ω
Current through each resistor = `"e"/"R"_"s" = 10/102.5 = 0.096"A"`
Power dissipated in each resistor =I2R (0.096)2 (50) = 0.46 watt
(b) When R1 and R2 are connected in parallel, voltage across each resistor is same and hence same power is dissipated in each resistor.
Total resistance of the circuit Rp = `(1/"R"_1 + 1/"R"_2)^-1 + r = 25 + 2.5 = 27.5 Omega`
Current through the circuit, I= `"e"/"R"_"p" = 10/27.5 = 0.36 "A"`
Voltage across each resistor, V = e -Ir= 10 - ( 0.36 x 2.5) = 9.1V
Power dissipated in each resistor P = `"V"^2/"R" = (9.1)^2/50 = 4140.5 "watt"`
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