Q11: An electric iron draws a current of 3.4 A from the 220 V supply line. What current will
this electric iron draw when connected to 110 V supply line?
Q12: A 6 ohm resistance wire is doubled up by folding. Calculate the new resistance of the wire.
Answers
Answer 11.
Given that, an electric iron draws a current of 3.4 A from the 220 V supply line.
We have to find the current of the electric iron drawn when connected to 110 V supply line.
From ohm's law we can say that,
V = IR
220 = 3.4 × R
220/3.4 = R
64.7 = R
So, the resistance is 64.7 ohm.
When electric iron is connected to 110 V supply line, then
V = 110V and R = 64.7 ohm
I = V/R
I = 110/64.7
I = 1.7 A
Therefore, the current is 1.7 A when connected to 110 V supply line.
Answer 12.
A 6 ohm resistance wire is doubled up by folding. (R = 6 ohm)
We know that,
R = p l/A = 6
Wire is doubled, means area is doubled and length becomes half the original length.
A = 2A and l = l/2
R' = p (l/2)/2A
R' = p l/4A
R' = 1/4 p l/A
R' = 1/4 × R
As, R = 6 ohm. So,
R' = 6/4
R' = 1.5
Therefore, the new resistance of the wire is 1.5 ohm.
Answer:
Explanation:Given that, an electric iron draws a current of 3.4 A from the 220 V supply line.
We have to find the current of the electric iron drawn when connected to 110 V supply line.
From ohm's law we can say that,
V = IR
220 = 3.4 × R
220/3.4 = R
64.7 = R
So, the resistance is 64.7 ohm.
When electric iron is connected to 110 V supply line, then
V = 110V and R = 64.7 ohm
I = V/R
I = 110/64.7
I = 1.7 A
Therefore, the current is 1.7 A when connected to 110 V supply line.
Answer 12.
A 6 ohm resistance wire is doubled up by folding. (R = 6 ohm)
We know that,
R = p l/A = 6
Wire is doubled, means area is doubled and length becomes half the original length.
A = 2A and l = l/2
R' = p (l/2)/2A
R' = p l/4A
R' = 1/4 p l/A
R' = 1/4 × R
As, R = 6 ohm. So,
R' = 6/4
R' = 1.5
Therefore, the new resistance of the wire is 1.5 ohm.