how to solve...
a battery of electromotive force of 2V and internal resistance 1 (ohm), is connected in series with a resistance of 15(ohm) & an ammeter of resistance 4(ohm). Find the reading of ammeter & the potential difference btwn the terminals.
ans:0.1 A, 1.9 V
2006-10-03 21:44:53 · 5 answers · asked by slimy dude 2 in Science & Mathematics ➔ Physics
first of all you calculte the total resistance.
Rtot= Rsource+ R +Rammeter = 1+15+4= 20 Ohms
so
I = V / Rtot = 2 / 20 = 0.10A
now we know that the loop is run with a current of 0.105A and the resistance of the ammeter is 4 ohms
therefore the terminal voltage is
Uterm = 0.1 * (R+Rammeter) =>
Uterm = 0.1 * 19 = 1.9V
take care
2006-10-03 22:23:34 · answer #1 · answered by Emmanuel P 3 · 0⤊ 0⤋
E = I(Ri+Ra+Rl) = I(1+4+15) = 2
I = 2/(1+4+15)
V = E - IRi
V = 2 - 0.1*1
2006-10-04 04:52:03 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
Ohm's law. voltage = current x resistance.
Add the resistance value, divide to get the current, apply law with that current across the resistance of the meter...
2006-10-04 04:50:56 · answer #3 · answered by Ken H 4 · 0⤊ 0⤋
well .1 A and 1.9 V r ryt, V=IR and proceed frm there nerdo
2006-10-04 04:54:36 · answer #4 · answered by sniper_llf 1 · 0⤊ 0⤋
Thanks.
2006-10-04 04:52:17 · answer #5 · answered by Anonymous · 0⤊ 0⤋