Question

Q.64 If a radio and a bulb each of resistance 3 ohm are connected in series to a 12
battery, the potential difference across each will be:
A) 3V
C) 9V
B) 6V
D) 12V

Answer 1

As we know, in a series connection, the current remains the same.

We know for a series connection:

$\boxed{\bold{\sf{R_{equivalent}=R_1+R_2 }}}$

Here, R₁ = R₂ = 3Ω

Applying the values of the equation:

$\sf{\implies R_{eq}=3+3 }$

$\sf{\implies R_{eq}=6 \ \Omega }$

We know the equation:

$\boxed{\bold{\sf{Current=\dfrac{emf}{Resistance}}}}$

Here,

• emf (E) = 12 E

• Resistance (R) = 3 Ω

Applying the values to the equation:

$\sf{\implies I=\dfrac{12}{6}}$

$\sf{\implies I=2 \ A}$

We know the equation:

$\boxed{\bold{\sf{Potential \ Difference, V=IR}}}$

Applying the values to the equation:

$\sf{\implies V=2\times3}$

$\sf{\implies V=6 \ V}$

$\huge{\boxed{\bold{\sf{\therefore Potential \ Difference=6 \ V}}}}$