Two equal resistances $R_{1}$ = $R_{2}$ = R are connected with a $30 Ω$ resistor and a battery of terminal voltage E. The currents in two branches are 2.25A and 1.5A as shown in fig. Then,

R_{2} = 15 Ω

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{2} = 60 Ω

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

E = 36 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

E = 180 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
B $R_{2} = 60 Ω$
D $E = 180 V$
A. Using Kirchhoff’s law, current through $R_{2} I S 2.25 - 1.5 = 0.75 A$ . Also, since $R_{2}$ is in parallel with the $30 Ω$
resistance, $R_{2}$ must be $60 Ω$ since only half the current flows through it compared to the current through $30 Ω$ resistor. Total resistance in circuit becomes, $(60 + 20) Ω = 80 Ω$ Potential drop across the battery, $E = I R = 2.25 \times 80 = 180 V$

Suggest Corrections

Similar questions

Q. Two ideal batteries of emf

V_{1}

and

V_{2}

and resistances

R_{1}

R_{2}

and

R_{3}

are connected as shown in the figure. The current in resistance

R_{2}

would be zero if

Q. Two resistors of resistance

R_{1}

and

R_{2}

having

R_{1} > R_{2}

are connected in parallel. For equivalent resistance R, the correct statement is

When resistors $R_{1}$ and $R_{2}$ are connected in series across a battery of potential difference $V,$ the power dissipated by the resistors are $P_{1}$ and $P_{2}$ respectively. When they are connected in parallel across a battery of potential difference $E,$ the power dissipated is $P_{3}$ and $P_{4}$ respectively. Which one of the following is true?