Question

A current of 500mA flows in a series circuit contains an electric lamp and a conductor of 10 ohm when connected to 6 V battery.find the resistance of the electric lamp

Answer 1

Answer:

2 ohm

Explanation:

Current=500mA = 0.5 A

because 1mA= 10^-3 A

Conductor's resistance=10 ohm

Voltage/ P.D.=6V

Now according to ohm's law

V=IR

R=V/I

R=6/0.5

=6×10/5

=6×2

=12 ohm

NOW

Resistance of Conductor=10ohm

Resistance found=12ohm

THEREFORE,

RESISTANCE OF ELECTRIC LAMP

=12-10=2 ohm

Answer 2

The resistance of the electric bulb is 2 ohms

Explanation:

Given that,

Current flowing in the circuit, $I=500\ mA=500\times 10^{-3}\ A=0.5\ A$

The resistance of the conductor, $R_c=10\ \Omega$

The equivalent resistance of the series combination of an electric lamp and a conductor is given by :

$R=R_c+R_l$

Voltage of the circuit, V = 6 V

Let R is the effective resistance. It is given by :

$R=\dfrac{V}{I}\\\\R=\dfrac{6}{0.5}\\\\R=12\ V$

In series combination, the current flowing through each circuit is same. The resistance of the lamp is given by :

$R_l=R-R_c\\\\R_l=12-10\\\\R_l=2\ \Omega$

So, the resistance of the electric bulb is 2 ohms. Hence, this is the required solution.

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