Question
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
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.
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
Related Questions
1. An electrical circuit consists of a battery with 2 cells,each with an EMF of 12 volts and an inte...
A battery X of emf 6volt and internal resistance 2(ohms) is connected in series with a battery Y of...
a battery of three cells in series each of emf 3volt an internal resistances of 0.5 ohms is connecte...
A battery of three cells in series ,each of e.m.f. 2v and internal resistance 0.5 ohms is connected...