Question

In an experiment on Ohm’s law, voltage drop across a resistor has been measured as V = (100 ± 5)

volt and current through the resistor has been measured as I = (10.0 ± 0.2) ampere. The permissible

error in calculation of R is (a) 5% (b) 10% (c) 7 % (d) 12%​

Answer 1

Answer:

ok na

Explanation:

see picok.

solution is there in pic.

Answer 2

Answer:

The permissible error in calculation of R is 7% i.e.option(c).

Explanation:

First, let's find the resistance of the resistor without any errors,

$R=\frac{V}{I}$              (1)

Where,

V=voltage across the resistor

I=current flowing through the resistor

R=resistance of the resistor

From the question we have,

$V=100\pm5$        (2)            (5 is the error in measurement)

$I=10\pm 0.2$          (3)            (0.2 is the error in measurement)

By substituting the required values in equation (1) we have;

$R=\frac{100}{10}=10\Omega$

Now the error in R is given as,

$\frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}$                  (4)

By putting the required values in (2) we get;

$\frac{\Delta R}{R}=\frac{5}{100}+\frac{0.2}{10}=0.05+0.02=0.07$          (5)

So, the resistance with error is represented as=$(10\pm0.07)\Omega$

The percentage error in R is given as,

$\frac{\Delta R}{R}\times 100=0.07\times=7\%$              (by using equation (5))

Hence, the permissible error in calculation of R is 7% i.e.option(c).