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.

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answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6