Asked by Emmanuel
A battery of three cells in series each of emf 2v and internal 0.5 ohms is connected to 2ohms resistor in series with a parallel combination of two 3ohms resistors (a)draw the circult diagram (b)the effective of external resistance (c)the current in the circult (d)the lost volt in the battery (e)the current in one of the 3ohms resistor
Answers
Answered by
Henry
I can't draw the circuit here, but i can analize the circuit:
a. Rt = 0.5 + 2 + (3 * 3) / (3 + 3) = 0.5 + 2 + 1.5 = 4 ohms = total
resistance in circuit.
b. The external circuit reduces the battery voltage.
c. I = V/Rt = 6 / 4 = 1.5Amps = the
current in circuit.
d. Vi = I*Ri = 1.5 * 0.5 = 0.75 volts
= Voltage drop across internal resistance.
e. The current divides equally between
the two 3 ohm resistors:
I3 = I / 2 = 1.5 / 2 = 0.75 Amps =
current through each 3 ohm resistor.
a. Rt = 0.5 + 2 + (3 * 3) / (3 + 3) = 0.5 + 2 + 1.5 = 4 ohms = total
resistance in circuit.
b. The external circuit reduces the battery voltage.
c. I = V/Rt = 6 / 4 = 1.5Amps = the
current in circuit.
d. Vi = I*Ri = 1.5 * 0.5 = 0.75 volts
= Voltage drop across internal resistance.
e. The current divides equally between
the two 3 ohm resistors:
I3 = I / 2 = 1.5 / 2 = 0.75 Amps =
current through each 3 ohm resistor.
Answered by
Emmanuel
Wow incredible nonsensical and then you are not sure of this questions you asked.
Answered by
Amira
Yes
Answered by
Timileyin
This is rubbish
Answered by
Ibukun
Good
Answered by
Onoja joy
It is nonsense
Answered by
Distinct
you did a great work.. but I. your calculation of effective external resistance, 2+1.5 = 3.5ohm and not 4ohm which affected further calculations..
Answered by
Simeon
Wow what rubbish
Answered by
Adebisi
a. Rt = 0.5 + 2 + (3 * 3) / (3 + 3) = 0.5 + 2 + 1.5 = 4 ohms = total
resistance in circuit.
b. The external circuit reduces the battery voltage.
c. I = V/Rt = 6 / 4 = 1.5Amps
d. Vi = I*Ri = 1.5 * 0.5 = 0.75
e. The current divides equally between
the two 3 ohm resistors:
I3 = I / 2 = 1.5 / 2 = 0.75 Amps
resistance in circuit.
b. The external circuit reduces the battery voltage.
c. I = V/Rt = 6 / 4 = 1.5Amps
d. Vi = I*Ri = 1.5 * 0.5 = 0.75
e. The current divides equally between
the two 3 ohm resistors:
I3 = I / 2 = 1.5 / 2 = 0.75 Amps
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