Question

Volt battery draw a current of 5 A. A bulb of 0.9 ohm is connected to it. Calculate the power dissipation of bulb​

Answer 1

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The total resistance of a circuit connected to the given cell of emf 15 V is the sum of all the resistances from various sources like, resistors, ammeter, voltmeter, etc., connected in series along with the internal resistance of the cell.

In this case, the total resistance is given as 3Ω+3Ω+6Ω=12ohms.

From the ohm's law the total current can be calculated from the formula I=V/R.

That is, I=V/R = 15/12 =V/A =1.25A.

Hence, the current through the battery is 1.25 amperes.

The voltage output of a device is measured across its terminals and is called its terminal voltage V. Terminal voltage is given by the equation V=emf−Ir where, r is the internal resistance and I is the current flowing at the time of the measurement.

Therefore, V=15V−(1.25×3)=11.25V.

Hence, the potential difference across the terminals of the cell is 11.25V