Home
Class 12
PHYSICS
In the circuit shown, the internal resis...

In the circuit shown, the internal resistances of the batteries are indicated next to them. Choose the correct option(s):

A

The potential at P is -7.5

B

The potential at Q is -1V

C

The potential at R is zero

D

The potential at S is zero

Text Solution

Verified by Experts

The current in the circuit `I=((20-10))/((7.5+0.5+1+1))=10/10-1amp`
The potential difference across PS, `V_(ps)=7.5 times 1=7.5V`
The potential difference across Q,R, `V_(QR)=1 times 1=1V`
As point G is connected to earth, hence potentials of R and S are zero.
The direction of the current in the circuit is from P to S, hence point P is at higher potential
`therefore V_p=7.5V`
Similarly, `V_Q=-V therefore` Hence (B) ( C) and (D) are correct
Promotional Banner

Topper's Solved these Questions

  • DC CIRCUIT

    VMC MODULES ENGLISH|Exercise LEVEL-1 JEE MAIN ARCHIVE|93 Videos

  • DC CIRCUIT

    VMC MODULES ENGLISH|Exercise JEE ADVANCED ARCHIVE|68 Videos

  • DC CIRCUIT

    VMC MODULES ENGLISH|Exercise LEVEL-2 DAILY TUTORIAL SHEET-4|10 Videos

  • CURRENT ELECTRICITY

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE-F|10 Videos

  • DYNAMICS OF A PARTICLE

    VMC MODULES ENGLISH|Exercise JEE Advance (Archive) Level - II (SINGLE OPTION CORRECT TYPE )|31 Videos

Similar Questions

Explore conceptually related problems

In the circuit shown,choose the correct option (All batteries are ideal)

For the resistance network shown in the figure, choose the correct options (s)

Shown n batteries connected to form a circuit. The resistances denote the internal resistances of the batteries which are related to the emf as r_(i)=k(epsilon)_(i) where K is a constant. The solid dots represent the terminals of the batteries. Find (a)the current through the circuit and (b) the potential difference between the terminals of the ith battery.

A capacitor of 4 mu F is connected as shown in the circuit. The internal resistance of the battery is 0.5 Omega . The amount of charge on the capacitor plates will be .

A capacitor of 4 mu F is connected as shown in the circuit. The internal resistance of the battery is 0.5 Omega . The amount of charge on the capacitor plates will be .

The reading of a galvanometer connected in the circuit is zero. If the internal resistance of a battery is zero then find the value of X.

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

A battery has a variable EMF E depending on the amount of electrochemical energy stored in it, U : E=E_0[V/V_0]^(1/3) , where E_0 is the EMF of the battery when it has the maximum possible energy, V_0 , stored in it. The internal resistance of the battery is negligible. The battery, with energy stored in it, is connected with a resistance R and the switch is closed at . Choose the correct option(s):

In the circuit shown, the resistances are given in ohms and the battery is assumed ideal with emf equal to 3.0 V.

Considering the graph of 1s , choose the correct option: