Electrical Engineering - Ohm's Law - Discussion
Discussion Forum : Ohm's Law - General Questions (Q.No. 13)
13.
The current through a flashlight bulb is 40 mA and the total battery voltage is 4.5 V. The resistance of the bulb is approximately
112 
11.2
1.2
18
Discussion:
10 comments Page 1 of 1.
Vvk said:
6 years ago
4.5/40 = .1125 unit is not changed.
Santhosh elli said:
7 years ago
R = V/I.
R = 4.5/40 * 10^-3,
R = 4.5 * 10^3/40.
R = 4500/40.
R = 450/4.
R = 112.5ohm.
R = 4.5/40 * 10^-3,
R = 4.5 * 10^3/40.
R = 4500/40.
R = 450/4.
R = 112.5ohm.
(1)
Rahul singh said:
7 years ago
R = V/I.
R = 4.5/40 * 10^-3,
R = 4.5 * 10^3/40.
R = 4500/40.
R = 450/4.
R = 112.5ohm.
R = 4.5/40 * 10^-3,
R = 4.5 * 10^3/40.
R = 4500/40.
R = 450/4.
R = 112.5ohm.
Sham said:
7 years ago
I = V/R.
R = V/A.
R = 4.5/40x10^-3.
R = 4.5x10^3/40.
R = 4500/40.
= 450/4.
= 112.2ohm.
R = V/A.
R = 4.5/40x10^-3.
R = 4.5x10^3/40.
R = 4500/40.
= 450/4.
= 112.2ohm.
Mamatha paladugu said:
7 years ago
V = IR.
R = V/I.
R = 4.5/40*10^-3.
=112.5ohms.
R = V/I.
R = 4.5/40*10^-3.
=112.5ohms.
Sharanabasava said:
1 decade ago
I = 40mA = 40*10^-3.
V = 4.5v.
Then R = V/I.
= 4.5/(40*10^-3).
= 4.5*10^3/40.
= 112.5 ohms.
V = 4.5v.
Then R = V/I.
= 4.5/(40*10^-3).
= 4.5*10^3/40.
= 112.5 ohms.
Arun said:
1 decade ago
V = IR.
R = V/I.
R = 4.5/0.04.
R = 112.5 Ohm.
R = V/I.
R = 4.5/0.04.
R = 112.5 Ohm.
Rajib said:
1 decade ago
V = IR,
R = V/I,
R = 4.5/40(mA).
R = 4.5/(40/1000)A.
R = 4.5/0.04.
R = 112.5 ohm.
R = V/I,
R = 4.5/40(mA).
R = 4.5/(40/1000)A.
R = 4.5/0.04.
R = 112.5 ohm.
RENUGADEVI said:
1 decade ago
R=V/I
=4.5/40*10^-3
=111.65
=4.5/40*10^-3
=111.65
Raghu said:
1 decade ago
R=V/I
4.5/.040
=112.5
4.5/.040
=112.5
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