Solveeit Logo

Question

Question: Two flash light electric incandescent lamps each requiring \(3 amp\) at \(1.5V\) are placed in serie...

Two flash light electric incandescent lamps each requiring 3amp3 amp at 1.5V1.5V are placed in series and connected to a 6V6V cell. What resistance must be connected in series to operate them?
A. 22Ω B. 2Ω C. 6Ω D. 1Ω \begin{aligned} & A.\text{ }22\Omega \\\ & B.\text{ }2\Omega \\\ & C.\text{ }6\Omega \\\ & D.\text{ }1\Omega \\\ \end{aligned}

Explanation

Solution

In order to find resistance that is connected in series with lamps we have to find the resistance of two incandescent lamps by using ohm’s law after that we will find resistance by putting a series of resistance formulas.

Formula used:
R=VIR=\dfrac{V}{I}
And,
Req=R1+R2+R3{{R}_{eq}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}

Complete step by step solution:
It is given that in the question that each lamp required 1.5V and 3A current
\to If the resistance of two lamps is R1 and R2{{R}_{1}}\text{ }and\text{ }{{R}_{2}} then,
R1=R2=VI =1.53 R1=R2=0.5Ω.....(1) \begin{aligned} & {{R}_{1}}={{R}_{2}}=\dfrac{V}{I} \\\ & =\dfrac{1.5}{3} \\\ & {{R}_{1}}={{R}_{2}}=0.5\Omega .....\left( 1 \right) \\\ \end{aligned}
Where R = resistance
V = voltage
I = current

\to Now we know that both the lamps are connected in series with the resistance R hence the total equivalent resistance is given by
\to Req=R1+R2+R.....(2){{R}_{eq}}={{R}_{1}}+{{R}_{2}}+R.....\left( 2 \right)
Req{{R}_{eq}} = total resistance in the series
R1,R2,R3{{R}_{1}},{{R}_{2}},{{R}_{3}} = resistance

Now substitute value of equation (1) in equation (2)
Req=0.5+0.5+R Req=1+R \begin{aligned} & \Rightarrow {{R}_{eq}}=0.5+0.5+R \\\ & \therefore {{R}_{eq}}=1+R \\\ \end{aligned}

\to Now according to ohm’s law we can use the below formula to find resistance R.
I=VReq....(3)I=\dfrac{V}{{{R}_{eq}}}....\left( 3 \right)
\to Here we will take V=6VV=6V which is given in the question.

Now substitute all the values in equation (3)
3=61+R 3+3R=6 3R=3 R=1Ω \begin{aligned} & \Rightarrow 3=\dfrac{6}{1+R} \\\ & \Rightarrow 3+3R=6 \\\ & \Rightarrow 3R=3 \\\ & \therefore R=1\Omega \\\ \end{aligned}

\to Hence the required resistance to operate incandescent lamps is R=1ΩR=1\Omega . Hence the correct option is (D).

Additional information:
When an electric current flows through a bulb or any conductor, the conductor offers some obstruction to the current and this obstruction is known as electrical resistance and is denoted by R.
\to Factors Affecting Resistance:
\to The temperature of the conducting material
\to Length of the conductor
\to The cross-sectional area of the conductor
\to The material of the conductor

Note:
When we are finding the resistance for a separate lamp we have to take voltage 1.5V but when we are calculating for all 3 resistances we have to take 6V cell voltage.