- Introduction
- |
- ElectroMotive Force(emf)
- |
- Internal Resistance of Battery (or cell)
- |
- Electric Energy and Power
- |
- Kirchoff's Rules
- |
- The junction Rule (or point rule)
- |
- The Loop Rule (or Kirchoff's Voltage Law)
- |
- Grouping of the cell's
- |
- Meterbridge (slide wire bridge)
- |
- Potentiometer
- |
- Comparison of EMF's of two cells using potentiometer
- |
- Determination of internal resistance of the cell

If current I enter at point A and leaves at diagonally opposite point G then the equivalent resistance of the cube would be

- 12R/5
- 12R
- 6R/5
- 5R/6

Consider the figure given below

Here AB is a wire of total resistance R

Current through ideal ammeter for any 0<p<1 is

Value of p for which ammeter reads maximum value is

(a) 0

(b) 1

(c) 2

(d) 3

Consider the circuit given below

Total resistance in the circuit is

(a) 5Ω

(b) 6Ω

(c) (11/6)Ω

(d) 3Ω

Current in the circuit is

(a) 4A

(b) 12A

(c) 6A

(d) 2A

Potential difference between points A and E is

(a) 12V

(b) 6V

(c) 10V

(d) 5V

Potential at point E is

(a) 10 V above the ground

(b) 10 V below the ground

(c) 2 V above the ground

(d) 2 V below the ground

Power supplied by the battery is

(a) 12 W

(b) 10W

(c)22W

(d)24W

Consider the two circuits given below

Column A |
Column B |

(P) Point of lowest potential in circuit A has its value equal to |
(U) 0.15 A |

(Q) Circuit current in circuit B is |
(V) 4V |

(R) Circuit current in circuit A is |
(W) 15V |

(S) Potential of point B in circuit B is |
(X) 90V |

(T) Potential at point B in circuit A is |
(Y) .5A |

In the circuit shown below the reading of ammeter is the same when both switches open as both switches close.

The value of resistance R is

- 800Ω
- 900Ω
- 1200Ω
- 1000Ω

For questions from 11 to 13 consider the statement given below

A 20µF capacitor which is initially uncharged is connected to a 6V battery through a resistor with resistance equal to 200Ω

Magnitude of final charge q

How long would it take for capacitor to be charged to (1/2)q

(a) 3.8×10

(b) 2.8×10

(c) 2.8×10

(d) 3.8×10

The time taken by the capacitor to be charged to 0.80q

In the figure given below the circuit contains four capacitors having same capacitance of 10µF and a battery providing a voltage of 90V

Match the columns given below

Column A |
Column B |

(P) When switch S_{2} is open and S_{1} is closed and then opened after C_{1} , C_{2} and C_{3} are fully charged. Electric potential difference across each capacitor is |
(W) 0 V |

(Q) Now after opening S_{1} if switch S_{2} is closed, the electric potential difference across each capacitor is |
(X) V_{1}=V_{3}=36V , V_{2}=V_{4}=18V |

(R) In another case if switch S_{1} is open and switch S_{2} is first closed then electric potential difference across each capacitor is |
(Y) V=30 V |

(S) Now if switch S_{1} is closed , the potential difference across each capacitor is |
(z) V_{1}=V_{3}=30V , V_{2}=V_{4}=15V |

- (d)
- (b)
- (a) , (b)
- (b)
- (d)
- (c)
- (c)
- (d)
- P—W ; Q—Y ; R—U ; S—V ; T—X
- (c)
- (a)
- (b)
- (c)
- P—Y ; Q—Z ; R—W ; S—X

Class 12 Maths Home page Class 12 Physics Home page

- NCERT Solutions: Physics 12th
- NCERT Exemplar Problems: Solutions Physics Class 12
- Concepts of Physics - Vol. 2 HC Verma
- Dinesh New Millennium Physics Class -XII (Set of 2 Vols) (Free Complete Solutions to NCERT Textbook Problems & NCERT Exemplar Problems in Physics-XII)
- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- CBSE Chapterwise Questions-Solutions Physics, Class 12
- CBSE 15 Sample Question Paper: Physics for Class 12th

- ncert solutions for class 6 Science
- ncert solutions for class 6 Maths
- ncert solutions for class 7 Science
- ncert solutions for class 7 Maths
- ncert solutions for class 8 Science
- ncert solutions for class 8 Maths
- ncert solutions for class 9 Science
- ncert solutions for class 9 Maths
- ncert solutions for class 10 Science
- ncert solutions for class 10 Maths