Question

plz answer the question ​

Answer 1

Answer:

The resistors are in series connection

(I) the current through 5 ohm is

V= IR

10 = I x 5

I =2 A

(ii) the current through R is also 2A bcoz this is series connection

(iii) V =IR

6 =2 X R

R=3 Ohm

(iv) V =IR

Total resistance = 3 +5 =8 ohm

V = 2x 8

V =16v

Answer 2

(a)In the Figure, there are 2 resistance $R_{1}$ and $R_{2}$ are connected in series with Battery V volts.

Let p.d. across $R_{1}$ is $V_{1}$ and p.d. across $R_{2}$ is $V_{2}$.

s.t. : V= $V_{1}$ $+$ $V_{2}$            ................{1}

Let, the equivalent resistance be R and current flowing be I.

By Ohm's Law,

                               $\frac{V}{I}=R$

                                        V = I × R                                      ................{2}    

Applying Ohm's law to both $R_{1}$ and $R_{2}$

                                     V1 = I × R1                                     ................{3}

                                     V2 = I × R2                                   ................{4}

From, Equation (1)(2)(3)(4)

I × R = (I × R1) + (I × R2)

I × R = I × (R1 + R2)

R = R1 + R2

(b)Following:

1. Current through 5 ohm resister = 2A.
2. Current through R = 2A.

    3. V = IR

        R = V/I = 6/2 = 3 ohm

   4. 10+6 = 16V