An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force(emf) 9.0V and negligible internal resistance as shown in fig 6.1 Wire X has a resistance of 2.4 ohm and wire Y has a resistance of 1.2ohm. A voltmeter is connected in parallel with the wires. A variable resistor is used to adjust the power dissipated in wires X and Y.

The variable resistor is adjusted so that the voltmeter reads 6.0V.

Calculate the resistance of the variable resistor.

0.4 ohms.

Explanation:

From the circuit,

The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.

1/R' = 1/R1+1/R2

R' = (R1×R2)/(R1+R2)

R' = (2.4×1.2)/(2.4+1.2)

R' = 2.88/3.6

R' = 0.8 ohms.

Hence the current flowing through the circuit is

I = V'/R' Equation 1

Where V' = voltmeter reading

I = 6/0.8

I = 7.5 A

This is the same current that flows through the variable resistor.

Voltage drop across the variable resistor = 9-6 = 3 V

Therefore, the resistance of the variable resistor = 3/7.5

Resistance = 0.4 ohms.

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

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