Question

The supply voltage to rom is 120V. The resistance of the lead wires is $6\mathrm{\Omega }$. A 60W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240W heater is switched on in parallel to the bulb?
(A) Zero
(B) 2.9 volt
(C) 13.3 volt
(D) 10.04 volt

Answer 1

Hint: Resistance is a measure of the opposition to the current flow in an electrical circuit. It is measured in ohms, and is symbolized by the Greek letter omega. When a voltage is applied across the substance there will be an electric current through it. Resistance causes some of the electrical energy to turn into heat so some electrical energy is lost along the way. Therefore, it is sometimes useful to add components called resistors into an electrical circuit to restrict the flow of electricity and protect the components in the circuit.

Complete step-by-step answer:
Resistance of the bulb is given to us as$R_b$
Using the formula, $P={\displaystyle \frac{V^2}{R}}orR={\displaystyle \frac{V^2}{P}}$
Now,
$Rb={\displaystyle \frac{{120}^2}{60}}=240\mathrm{\Omega }$
Similarly, for the heater the resistance is given as,
$R_n={\displaystyle \frac{{120}^2}{240}}=60\mathrm{\Omega }$
Now, the equivalent resistance of the bulb and heater together is given as:
$R={\displaystyle \frac{R_b{R}_n}{R_b+R_n}}={\displaystyle \frac{240\times 60}{240+60}}=48\mathrm{\Omega }$
Before the heater was connected, the voltage drops across the bulbs
$V_2={\displaystyle \frac{12}{R_b+6}}\times R_b={\displaystyle \frac{120}{240+6}}\times 240=117V$
After the heater is connected, the voltage is dropped by
$V_1={\displaystyle \frac{120}{R+6}}\times R={\displaystyle \frac{120}{48+6}}\times 48=106.66V$
So, $V_2-V_1=10.04V$

Hence, the correct answer is Option D.

Note: The equivalent resistance of a network is that single resistor that we can use to replace the entire network in such a way that for a certain applied voltage V we will get the same current I as we were getting for a network. If the two resistances or impedances in parallel are equal and of the same value, then the total or equivalent resistance is given as the equal to half the value of one resistor.
It should be known to us that when resistors are connected in parallel, more current flows from the source than would flow for any of them individually, so the total resistance is lower. Each resistor in parallel has the same full voltage of the source applied to it, but divide the total current amongst them.