Electric Circuit - 1

Since $\displaystyle R_1$ and $R_2$ are in parallel,

$V_1=V_2 \\ \Rightarrow I_1R_1 = I_2R_2 \\ \Rightarrow 4 A \times 30 Ω = I_2 \times 60 Ω \\ \Rightarrow I_2 = 2 A$

From Kirchoff's First Law, $I_3 = I_1 + I_2 = 4A + 2A = 6A$

$V_3 = I_3R_3 = 6A \times 70 Ω = 420 V$

Now, $V = V_1 + V_3 = 120V + 420V = \boxed{540V}$

@Fahim Shahriar Shakkhor your's is definitely easier but just see how a I did it.

I 'll get through this problem with some other approach.

First of all the total $R=90$ .

For those who want to know how :

As $R_1$ and $\displaystyle{R_2}$ are parallel so $\displaystyle{\frac{1}{R_p}=\frac{1}{R_1} + \frac{1}{R_2}}$ where $R_p$ is the net resistance of the parallel circuits, it comes out to be, $\displaystyle{R_p}=20$ and its seen that $R_3=70$ and hence we arrive at the total $R$ .

The current flowing through the wire $I$ Now, the current in a parallel circuit get divided in the ratio of the individual resistances, i.e.. in $\displaystyle{R_1}$ the current flow 'd be $\dfrac{60}{90} \times I$ and in $R_2$ the current flow'd be $\dfrac{30}{90} \times I$ , In $\displaystyle{R_1}$ it is given $4$ so plugging the values in, we get,

Note : That the current flow in $R_1$ is more than in $\displaystyle{R_2}$ because $\displaystyle{R_1 \lneq R_2}$ and this is given by the ratios.

$I=6$

$V=IR$ ....and.. $V=\boxed{540}$

Let the currents through $R_1$ , $R_2$ and $R_3$ be $I_1$ , $I_2$ and $I_3$ respectively, voltage across $R_1$ and $R_2$ be $V_1$ and that across $R_3$ be $V_2$ .

Then we have:

$I_1 = 4A \quad \Rightarrow V_1 = I_1R_1 = 4\times 30 = 120V$

$I_2 = \dfrac {V_1}{R_2} = \dfrac {120}{60} = 2A$

$\Rightarrow I_3 = I_1+I_2 = 4+2 = 6A\quad \Rightarrow V_2 = I_3R_3 = 6\times 70 = 420V$

$\Rightarrow V = V_1+V_2 = 120 + 420 = \boxed {540} V$

It is given that current through R1(30 ohm) is 4A, hence current through R2 (60 ohm) must be half of current through R1 .Hence current through R2 is 2A. So net current supplied by battery is equal to 6A.

NOW parallel combination of R1 and R2 is in series with R3, hence equivalent resistance of given circuit is equal to 90 ohm. applying ohm's law;

V=IR; V=net current supplied by battery* equivalent resistance .

V=6*90=540 ohm.

potential difference of R1 or R2 is = 30 ohm * 4A = 120V . But the equivalent resistance of R1 & R2 , is Rp = 1/(1/30 + 1/60) ohm = 20 ohm. So, the current flowing through the series connection of Rp & R3 is = 120V/20ohm = 6A . But the equivalent resistance of the whole circuit is = Rp + R3 = 70 ohm + 20 ohm = 90 ohm. So the Voltage in this circuit is = 90ohm * 6A = 540V.

use current division rule