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 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
9 answers
Wow incredible nonsensical and then you are not sure of this questions you asked.
Yes
This is rubbish
Good
It is nonsense
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..
Wow what rubbish
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
0 answers