Complete question is;
A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?
Answer:
150.42 V
Explanation:
We are told that the circuit consists of four 100W lamps.
We know that Power is given by the equation;
P = V²/R
Thus;
R = V²/P
Now, we are told that the four lamps are connected in parallel across a 230V supply.
Thus, V = 230 V
So resistance, R = 230²/100
R = 529 Ω
We are told that the resistance of the lamps under the conditions stated is six times their value when burning normally.
Thus, total resistance of each lamp under the conditions = 529 × 6 = 3174 Ω
So, since they are connected in parallel, equivalent resistance for each lamp = 3174/4 = 793.5 Ω
Now, since this resistance is connected in series with the voltmeter resistance of 1500 Ω
Therefore, total circuit resistance = 1500 + 793.5 = 2293.5 Ω
Thus;
circuit current = 230/2293.5 = 0.100283 A
Now, according to Ohm’s law, voltage drop across the voltmeter = 1500 × 0.100283 ≈ 150.42V
