Question

three resistance of 4 ohm connected with battery of 3v . calculate the current drawn by the battery​

Answer 1

GiveN :-

• Three resistors of 4Ω are Connected in parallel to a resistance of 4Ω .
• The voltage of the battery is 3V .

To FinD :-

• The current drawn out of the cell .

SolutioN :-

Here since the three 4 Ω resistance are in series hence the net resistance will be 4Ω×3 = 12Ω . Now this is connected to 4Ω in parallel . So the net resistance will be ,

$\sf:\implies \pink{\dfrac{1}{R_{(net)}}=\dfrac{1}{R_1}+\dfrac{1}{R_2} }\\\\\sf:\implies R_{(net)}= \dfrac{R_1R_2}{R_1+R_2}\\\\\sf:\implies R_{(net)}=\dfrac{(12\Omega )(4\Omega)}{(12+4)\Omega}\\\\\sf:\implies R_{(net)} = \dfrac{48\Omega^2}{16\Omega}\\\\\sf:\implies \boxed{\pink{\frak{ R_{(net)} = 3\Omega }}}$

$\rule{200}2$

☯️ $\underline{\boldsymbol{ Using \ Ohm's \ Law :- }}$

$\sf:\implies\pink{ Voltage = ( Current)(Resistance)}\\\\\sf:\implies 3V = I \times 3\Omega \\\\\sf:\implies I =\dfrac{3}{3} A \\\\\sf:\implies \underset{\blue{\sf Required \ Current }}{\underbrace{\boxed{\pink{\frak{ Current = 1 \ Ampere }}}}}$