MCQ
 In the circuit shown below, calculate the current flowing through 20V cell.
Image
  • A
    12A
  • B
    15A
  • C
    12.5A
  • D
    12.3A

Answer

  1. 12.5A

Explanation:

Here, we can apply Kirchhoff’s loop rule in closed loops ADCBA, AEFBA, AHGBA, and AIJBA. Thus, we get

20 + 10 – 5I1 = 0 → I1 = 6A

-5 + 20 – 15I2 = 0 → I2 = 1A

25 + 20 – 10I3 = 0 → I3 = 4.5A

-15 + 20 – 5I4 = 0 → I4 = 1A

Therefore, the current flowing through the 20V cell = 6 + 1 + 4.5 + 1 = 12.5A

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

A plane mirror reflecting a ray of incident light is rotated through an angle θ about an axis through the point of incidence in the plane of the mirror perpendicular to the plane of incidence, then

(a) The reflected ray rotates through an angle 2θ

(b) The incident ray is fixed

(c) The reflected ray does not rotate

(d) a, b

The wavelength of light visible to eye is of the order of

(a)  

(b)  m

(c) 1 m    

(d) 6  m

Cutoff wavelength of X-rays coming from a Coolidge tube depends on the:

Solar energy is mainly caused due to

(a) Fission of uranium present in the sun

(b) Fusion of protons during synthesis of heavier elements

(c) Gravitational contraction

(d) Burning of hydrogen in the oxygen

Monochromatic light of frequency 5  Hz travelling in vacuum enters a medium of refractive index 1.5. Its wavelength in the medium is

(a) 4000 Å

(b) 5000 Å

(c) 6000 Å

(d) 5500 Å

Two point charges Q  and – 3Q are placed at some distance apart. If the electric field at the location of Q is E then at the locality of -3Q, it is

(a) - E

(b) E/3

(c) -3E

(d) -E/3

The rms speed of oxygen molecules in a gas is v. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become:
  1. $\text{v}$
  2. $\text{v}\sqrt{2}$
  3. $2\text{v}$
  4. $4\text{v}$

A charge produces an electric field of 1 N/C at a point distant 0.1 m  from it. The magnitude of charge is

(a) 1.11  

(b) 9.11  

(c) 7.11  

(d) None of these

The inverse square law of intensity $\Big(\text{i.e., the intensity}\propto\frac{1}{\text{r}^2}\Big)$ is valid for a:
  1. Point source.
  2. Line source.
  3. Plane source.
  4. Cylindrical source.
A lightning arrester must have the following property.