Question

Two resistors of resistances 2 Ω and 4 Ω are connected in series with a battery of 6 V with an ammeter to measure the current through 4 Ω resistor and a voltmeter to measure the voltage across 2 Ω resistor with a closed plug key. (i) Draw the circuit diagram. (ii) Find readings of ammeter and voltmeter.

Answer 1

Answer:

Energy dissipated = PtorVI×tor =I

2

Rt Joules

Consider Ohm's law V=IR.

In this case the resistors are connected in series and hence, the total resistance is 4+2=6 ohms. In a series circuit, the current remains the same throughout the circuit and so we use the I

2

Rt Joules formula.

The total current in the circuit is calculated as I=V/R=6/6=1ampere.

Therefore, the power dissipated P across the 4 ohm resistor for 5 s

= 1

2

×4×5=20Joules

Hence, The heat dissipated by the 4 resistor in 5 s will be 20 J.

Answer 2

Answer:

Explanation:

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