Question

2 ohm and 4 ohm resistors are connected in parallel. 12 V potential difference is applied. Find the current in the circuit.

Answer 1

Answer:

Given ,

2Ω and 4Ω resistances are connected in parallel.

12 V potential difference is applied to those resistances .

We need to find the current in the circuit .

When resistances are connected in parallel . Then the resultant resistance is given by ,

\bf R_{parallel.Eq}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+.....R

parallel.Eq

=

R

1

1

+

R

2

1

+

R

3

1

+.....

Given that ,

\bf R_1=2\;\Omega\ ,\;R_2=4\;\OmegaR

1

=2Ω ,R

2

=4Ω

$$\begin{lgathered}\implies \bf \dfrac{1}{R_{P.Eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\implies \bf \dfrac{1}{R_{P.Eq}}=\dfrac{1}{2}+\dfrac{1}{4}\\\\\implies \bf R_{P.Eq}=\dfrac{4*2}{4+2}\\\\\implies \bf R_{P.Eq}=\dfrac{8}{6}\\\\\implies \bf R_{P.Eq}=\frac{4}{3}\;\Omega\end{lgathered}$$

So we have ,

Resistance , R = 4/3 Ω ,

Potential Difference , V = 12 V

Current , I = ? A

Now use Ohm's law to find the current .

$$\begin{lgathered}\implies \bf V=IR\\\implies \bf 12=I*\dfrac{4}{3}\\\\\implies \bf I=9\;Amperes\end{lgathered}$$

So current in circuit is 9 Amperes

Answer 2

Answer:

net resistance= 1/2 + 1/4

=3/4

by ohm's law

V=IR

I=R/V

3/4*12=9

HENCE, CURRENT IS 9 AMPERES