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
Yes
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..
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
Wow incredible nonsensical and then you are not sure of this questions you asked.
This is rubbish
Good
It is nonsense
Wow what rubbish
To solve this problem, let's break it down step by step:
(a) Draw the circuit diagram:
The given circuit consists of a battery of three cells in series, each with an emf (electromotive force) of 2V and an internal resistance of 0.5 ohms. This battery is connected to a 2-ohm resistor in series with a parallel combination of two 3-ohm resistors.
The circuit diagram would look like this:
________________________
| | |
+ 2V - 0.5Ω + 2V - 0.5Ω + 2V - 0.5Ω +
|___________|__________|
| | |
2Ω 3Ω 3Ω
(b) Calculate the effective external resistance:
The effective external resistance is the sum of all the resistances outside the battery (excluding the internal resistances of the cells). In this case, the external resistance is the 2-ohm resistor and the two 3-ohm resistors in parallel. To calculate the effective resistance, we need to find their equivalent resistance.
To find the equivalent resistance for the two 3-ohm resistors in parallel, we use the formula:
1/Req = 1/R1 + 1/R2
R1 = R2 = 3Ω
1/Req = 1/3Ω + 1/3Ω
1/Req = 2/3Ω
Req = 3/2Ω = 1.5Ω
The effective external resistance is the sum of the 2-ohm resistor and the equivalent resistance of the parallel combination:
Reffective = 2Ω + 1.5Ω
Reffective = 3.5Ω
(c) Calculate the current in the circuit:
To find the current in the circuit, we can use Ohm's Law:
V = I * R
The voltage (V) across the external resistance is the emf of the battery since the internal resistances are in series with the cells:
V = 2V + 2V + 2V
V = 6V
Using Ohm's Law, we can now calculate the current (I):
I = V / Reffective
I = 6V / 3.5Ω
I ≈ 1.714 A
(d) Calculate the lost volt in the battery:
The lost voltage in the battery is the voltage drop across the internal resistances. Since the internal resistances are in series with the cells, the lost voltage is the sum of the voltage drops across each internal resistance:
Vlost = I * Rinternal
Vlost = 1.714 A * 0.5Ω (since each cell has an internal resistance of 0.5Ω)
Vlost ≈ 0.857 V
(e) Calculate the current in one of the 3-ohm resistors:
Since the two 3-ohm resistors are in parallel, they have the same voltage drop. We can calculate the current through one of the 3-ohm resistors using Ohm's Law:
I3Ω = V / R
I3Ω = 6V / 3Ω
I3Ω = 2 A
So, the current in one of the 3-ohm resistors is 2 A.
That's it! You have successfully solved all the parts of the problem.
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.