A battery of e.m.f 15V and internal resistance 2 ohm is connected to two resistors of 4 ohm and 6 ohm joined in: (a) series (b) parallel In each case what is the electrical energy spent per minute...

A battery of e.m.f 15V and internal resistance 2 ohm is connected to two resistors of 4 ohm and 6 ohm joined in:

(a) series

(b) parallel

In each case what is the electrical energy spent per minute in 6 ohm resistor.

Asked on by novajose

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The emf of the battery is 15 V and the internal resistance is 2 ohm. The battery is connected to a circuit made of 2 resistors of 4 ohm and 6 ohm.

a. When the resistors are joined in series, the total resistance of the circuit is 4 + 6 + 2 = 12 ohm. The current passing through all the resistors is I = V/R = 15/12.

The current passing through the 6 ohm resistor is 1.25 A. The energy spent due to this is equal to power*60 = (1.25)^2*6*60 = 562.5 J

b. When the resistors are connected in parallel, their total resistance is (1/4 + 1/6)^-1 = 2.4 ohm. With the internal resistance, the total resistance of the circuit is 2 + 2.4 = 4.4 ohm. A voltage of 15 V causes a current of 15/4.4 = 75/22 A to flow through the circuit. This leads to a voltage of 90/11 V across the two resistors. The current through the 6 ohm resistor is 15/11 A.

The energy spent due to this in a minute is (15/11)^2*6*60 = 669.42 J.

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