Home
Class 12
PHYSICS
Consider a simple circuit shown in Fig. ...

Consider a simple circuit shown in Fig. 2(ET).2. stands for a variable resistance R'.R' can vary from `R_(0)` to infinity. r is internal resistance of the battery `(rltltRltltlR_(0))`

A

Potential drop across AB is nealry constant as `R^(')` is varied

B

Current through `R^(')` is nearly a constant as `R^(')` is varied

C

Current I depends sensitivity on `R^(')`

D

`Ige(V/(r+R))` always

Text Solution

Verified by Experts

The correct Answer is:
A, D
Promotional Banner

Topper's Solved these Questions

  • ELECTRIC CURRENT & CIRCUITS

    CENGAGE PHYSICS|Exercise Kirchhoff s Law and Grouping Cells|15 Videos
  • ELECTRIC CHARGES AND FIELDS

    CENGAGE PHYSICS|Exercise QUESTION BANK|40 Videos
  • ELECTRIC CURRENT AND CIRCUIT

    CENGAGE PHYSICS|Exercise Interger|8 Videos

Similar Questions

Explore conceptually related problems

For maximum power from battery the internal resistance of battery r is

For a circuit shown in fig find the value of resistance R_(2) and current I_(2) flowing thorugh R_(2)

Consider the circuit shown in Figure . The terminal voltage of the 24.0 V battery is 21.2V. What are (a) The internal resistance r of the battery and (b) The resistance R of the circuit resistor ?

Find the current flowing current flowing thorugh the resistance R in the circuit shown in Fig. The internal resistances of the batteries are negilble.

Consider the circuit shown in fig. Assuming R_1 = 2Omega, current passing through resistance R_1 is

In the circuit shows in Fig. 6.42 , resistors X and Y , each with resistance R , are connected to a 6 V battery of negligible internal resistance. A voltmetre, also of resistance R , is connected across Y . What is the reading of the voltmeter?

Consider circuit as shown . Which of the graph best represent V-l relationship across the variable resistance R.

A battery of emf epsilon and internal resistance r sends currents I_(1) and I_(2) , when connected to external resistance R_(1) and R_(2) respectively. Find the emf and internal resistance of the battery.

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) .