Question

A man has a body resistance of 600 . How much current flows through his ungrounded body . (i) when he touches the terminals of 12 v battery. (ii) when he sticks his finger into 120-v light socket.

Answer 1

Answer :-

I = 0.02 A

I = 0.2 A

Solution :-

Case 1 :-

$R = 600 \Omega$

V = 12 v

I = ?

According to ohm' s law :-

The magnitude of current is given by :-

$V = I R$

put the given value ,

→$12 = I \times 600$

→$l = \dfrac{12}{600}$

→$I = 0.02 A$

Case :- 2

$R= 600 \Omega$

V = 120

I = ?

→$120 = I \times 600$

→$I = \dfrac{120}{600}$

→$I = 0.2 A$

hence,

The magnitude of current in first case is 0.02 A and second case is 0.2 A.

Answer 2

Answer:

I = 0.02 A

I = 0.2 A

Explanation:

Given Problem:

A man has a body resistance of 600 . How much current flows through his ungrounded body . (i) when he touches the terminals of 12 v battery. (ii) when he sticks his finger into 120-v light socket.

Solution:

In,

Case 1:

$R = 600 \Omega$

$V =12v$

$I =? \quad (To Find)$

According to Ohm's law,

We know that,

The magnitude of current is given by,

$\bigstar V = IR\bigstar$

Now,

Put the values in equation:

$\implies\ 12 = I \times600$

$\implies\ l = \dfrac{12}{600}$

$\implies\ l =0.02A$

Now,

Case 2:

$R = 600\Omega$

$V = 120$

$I =?\quad (To Find)$

$120= I \times600$

$I = \dfrac{120}{600}$

$I = 0.2A$

Hence,

The magnitude of current in first case is 0.02 A and second case is 0.2 A