A battery of three cells in series ,each of e.m.f. 2v and internal resistance 0.5 ohms is connected to a 2 ohm resistor in series with a parallel combination of two 3 ohms resistor . Draw the circuit diagram and calculate:
a) the effective external resistance
b) the current in the circuit
c) the volts in battery
d) the current in one of the 3 ohm resistor
Effective resistance=R
Total resistance of circuit will be
R=2ohm +3×3\3+3ohm= (2+1.5)ohm=3.5
Equivalent resistance of the circuit Req= (3.5+1.5)=5ohm
Emf cell
V=E1+E2+E3=(2+2+2)v=6v
2. Current the circuit
=E/Req=6/5A
3. Lost voltage=v=Ir=6/5×1.5=1.8v
4. Current in the 3ohm resistor
(I1) =6/5×2=0.6A.
Hope it was helpful
Yes
Oh boy, time to get my drawing skills on! Here's a simple circuit diagram for your setup:
___ 2V ___ 0.5Ω ____ 2Ω ____ _____
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|____2V____ 0.5Ω _| |
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|____2V____ 0.5Ω_| |
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|___ | |
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|________________________|
Now, let's get into the calculations:
a) The effective external resistance is the sum of the resistors in parallel, which is (1/3 + 1/3)^-1 = 3/2 = 1.5Ω.
b) To find the current in the circuit, we can use Ohm's Law. The total resistance is the sum of the internal resistances and the external resistance. So, it's 1.5Ω + 0.5Ω + 0.5Ω = 2.5Ω. The current is given by I = V/R, where V is the total voltage and R is the total resistance. Plugging in the values, we get I = (2V)/(2.5Ω) = 0.8A.
c) The sum of the voltages in the battery is simply the sum of the individual voltages. So, it's 2V + 2V + 2V = 6V.
d) Since the circuit is series, the current passing through each 3Ω resistor will be the same as the current in the circuit, which is 0.8A.
I hope these calculations didn't shock you too much! Let me know if you have any more electrifying questions!
To draw the circuit diagram, follow these steps:
1. Place three cells in a series. Connect the positive terminal of the first cell to the negative terminal of the second cell, and the positive terminal of the second cell to the negative terminal of the third cell.
2. Assign each cell an electromotive force (e.m.f) of 2V and an internal resistance of 0.5 ohms.
3. Connect a 2 ohm resistor in series with the battery arrangement from step 1.
4. In parallel to the 2 ohm resistor, connect two 3 ohm resistors.
The resulting circuit diagram should look like this:
[2V, 0.5Ω]----[2V, 0.5Ω]----[2V, 0.5Ω]
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| 2Ω Resistor |
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----[3Ω]----[3Ω]-----
Now, let's calculate the given parameters:
a) Effective External Resistance:
To find the effective external resistance, we need to calculate the total resistance of the parallel combination of the two 3 ohm resistors.
The formula for the total resistance of two resistors in parallel is given by:
1/Re = 1/R1 + 1/R2
Substituting the given values:
1/Re = 1/3 + 1/3 = 2/3
Re = 3/2 = 1.5Ω
Hence, the effective external resistance is 1.5Ω.
b) Current in the Circuit:
To find the current in the circuit, we can use Ohm's Law:
I = V / R
The total voltage (V) across the external resistance can be calculated as follows:
V = Total e.m.f. - Total internal resistance
V = 2V + 2V + 2V - (0.5Ω + 0.5Ω + 0.5Ω) = 6V - 1.5Ω = 4.5V
Now, substituting the values in the formula:
I = V / Re = 4.5V / 1.5Ω = 3A
Therefore, the current in the circuit is 3A.
c) Volts in Battery:
The voltage across each cell will be its e.m.f minus its internal resistance. So, the voltage across each cell would be 2V - 0.5Ω = 1.5V.
Therefore, the volts in the battery are 1.5V.
d) Current in one of the 3 ohm resistors:
As the 3 ohm resistors are connected in parallel, they have the same voltage across them.
Using Ohm's Law:
I = V / R = 1.5V / 3Ω = 0.5A
Therefore, the current in one of the 3 ohm resistors is 0.5A.