Home
Class 12
PHYSICS
In the circuit shows in Fig. 6.63, the c...

In the circuit shows in Fig. 6.63, the cell is ideal with emf `9 V`. If the resistance of the coil of galvanometer is `1 Omega`, then

A

no current flows in the galvanometer

B

charge flowing through `8 mu F` is `40 mu C`

C

potential difference across `10 mu F` is `5 V`

D

potential difference across `10 mu F` is `4 V`

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
Since `R_(1)//R_(2) = C_(2)//C_(1)`, the wheatstone bridge is balanced.
Hence, `V_(C) = V_(D)`. No current passes through the galvanometer.
Hence, option (a) is correct.
Potential difference across `R_(1) =` potential difference across
`C_(1) = 4V`
Potential difference across `R_(2) =` potential difference across
`C_(2) = 5 V`
Potential difference across `5 Omega` is the potential difference across `8 mu F` capacitor. So, charge across `8 mu F` capacitor is
`Q = CV = 8 mu F xx 5V = 40 mu C`. hence, option (b) is correct and so is option(d).
Promotional Banner

Topper's Solved these Questions

  • ELECTRICAL MEASURING INSTRUMENTS

    CENGAGE PHYSICS|Exercise Assertion-Reasoning|7 Videos

  • ELECTRICAL MEASURING INSTRUMENTS

    CENGAGE PHYSICS|Exercise Comprehansion|15 Videos

  • ELECTRICAL MEASURING INSTRUMENTS

    CENGAGE PHYSICS|Exercise Single Correct|94 Videos

  • ELECTRIC POTENTIAL

    CENGAGE PHYSICS|Exercise DPP 3.5|14 Videos

  • ELECTROMAGENTIC INDUCTION

    CENGAGE PHYSICS|Exercise QUESTION BANK|40 Videos

Similar Questions

Explore conceptually related problems

In the circuit shown below the cell is ideal, with emf = 2 V: The resistance of the coil of the galvanometer G is 1 Omega . Then, find the potential difference across C_(1) (in volts), in steady state

In the circuit shown, the cell is ideal with emf = 2V . The resistance of the coil of the galvanometer G is 1 Omega . (i) No current flows in G (ii) 0.2A current flows in G (iii) potential difference across C_(1) is 1V (iv) Potential difference across C_(2) is 1.2V

In the circuit shown in fig. the cell is ideal with emf 15V. Each resistance is of 3Omega. The potential difference across the capacitoro in steady state is

In the circuit shown in fig. the cell has emf 10V and internal resistance 1 Omega

In the circuit shown, the cell is ideal , with emf= 15 V . Ecah resistance is of 3 Omega. the potential difference across the capacitor is

In the circuit shows in Fig. 6.74, the internal resistance of the cell is negligible. For the value of R = 40//x Omega , no current flows through the galvanometer. What is x ?

In the circuit shown in Fig. the emf of the sources is equal to xi = 5.0 V and the resistances are equal to R_(1) = 4.0 Omega and R_(2) = 6.0 Omega . The internal resistance of the source equals R = 1.10 Omega . Find the currents flowing through the resistances R_(1) and R_(2) .

In the electric circuit shown each cell has an emf of 2 V and internal resistance of 1Omega . The external resistance is 2Omega . The value of the current is (in A)

Draw the circuit diagram to verify Ohm's Law with the help of a main resistance of 100 Omega and tow galvanometer of resistance 10^6 Omega and 10^(-3)Omega and a source of varying emf. Show the correct positions of volmeter and ammeter.