Home
Class 12
PHYSICS
In the figure shown, what is the current...

In the figure shown, what is the current (in Ampere) drawn from the battery ? You are given:
`R_(1) =15 Omega,R_(2)=10 Omega,R_(3)=20Omega,R_(4)=5Omega`,
`R_(5)=25Omega,R_(6)=30Omega,E=15V`

A

`13//24`

B

`7//18`

C

`9//32`

D

`20//3`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • JEE MAINS

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise Chemistry|1 Videos
  • JEE MAIN

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise All Questions|452 Videos
  • JEE MAINS 2020

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise PHYSICS|250 Videos

Similar Questions

Explore conceptually related problems

The potential difference in volt across the resistance R_(3) in the circuit shown in figure, is (R_(1)=15Omega,R_(2)=15Omega, R_(3)=30Omega, R_(4)=35Omega)

A constant voltage V = 25 V is maintained between points A and B of the circuit (Fig). Find the magnitude and direction of the current flowing through the segment CD if the resistances are equal to R_(1) = 1.0 Omega, R_(2) = 2.0 Omega, R_(3) = 3.0 Omega and R_(4) = 4.0 Omega .

In the given cirucit, determine current through branch having indicated resistor R_(4) Given R_(1) = R_(2) = 4Omega, R_(3) = R_(4) = 2Omega R_(5) = 5Omega epsilon_(1) =5 V, epsilon _(2) =10V

Current through resistor R3 as shown in figure is ______. Given that R1=10 Omega R2=5 Omega R4=20 Omega and R5=10 Omega

Determine the voltage drop across the resistor R_1 in the circuit given below with E = 65V, R_1 = 50Omega, R_2 = 100 Omega, R_3 = 100 Omega and R_4 300 Omega

(a) Resistors given as R_(1), R_(2) and R_(3) are connected in series to a battery V. Draw the circuit diagram showing the arrangement. Derive an expression for the equivalent resistance of the combination. (b) If R_(1)=10 Omega, R_(2)=20 Omega and R_(3)=30 Omega , calculate the effective resistance when they are connected in series to a battery of 6 V. Also find the current flowing in the circuit.

In a given series LCR circuit R=4Omega, X_(L)=5Omega and X_(C)=8Omega , the current

Find the current through the resistance R in figure.If (a) R=12 Omega (b) R=48 Omega .

Find the current in each wire. epsilon = 50V, epsilon_2 = 40V epsilon_3 =30 V, epsilon_4 = 10V R_1 = 2Omega, R_2 = 2Omega R_3 = 1Omega

In the circuit shown below E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega and R_(3) = 2 Omega . The current I_(1) is