A 2.0-ohm resistor is connected in series with a 20.0-V battery and a three-branch parallel network with branches whose resistances are 8.0 ohms each. Ignoring the battery's internal resistance, what is the current in the battery?
First, we can find the equivalent resistance of the parallel network. The reciprocal of the total resistance in a parallel network is equal to the sum of the reciprocals of the individual resistances:
1/R_total = 1/R1 + 1/R2 + 1/R3
Since all three branches have the same resistance (8.0 ohms), we can rewrite this as:
1/R_total = 3*(1/8)
1/R_total = 3/8
So, the reciprocal of the total resistance is 3/8, which means the total resistance in the parallel network is:
R_total = 8/3 ohms (approximately 2.67 ohms)
Now, we can combine this resistance in series with the 2.0-ohm resistor:
R_combined = R_total + R_series = (8/3) + 2 = (8+6)/3 = 14/3 ohms (approximately 4.67 ohms)
Finally, we can find the current in the battery using Ohm's Law:
I = V/R = 20.0 V / (14/3) ohms ≈ 4.29 A
So, the current in the battery is approximately 4.29 amps.
To find the current in the battery, we need to calculate the equivalent resistance of the parallel network and then use Ohm's Law.
Step 1: Calculate the total resistance of the parallel network
The formula to calculate the total resistance of a parallel network is:
1/Req = 1/R1 + 1/R2 + 1/R3 + ...
In this case, since all three branches have the same resistance (8.0 ohms each), we can simplify the equation:
1/Req = 1/8.0 + 1/8.0 + 1/8.0
1/Req = 3/8.0
1/Req = 0.375
To find Req, we take the reciprocal of both sides of the equation:
Req = 1 / 0.375
Req = 2.6667 ohms (approximately)
Step 2: Calculate the current in the battery using Ohm's Law
Ohm's Law states that the current in a circuit is given by the equation:
I = V / R
In this case, the voltage is 20.0 V and the total resistance is 2.6667 ohms:
I = 20.0 / 2.6667
I ≈ 7.5 A
Therefore, the current in the battery is approximately 7.5 Amperes.
To find the current in the battery, we need to calculate the total resistance of the circuit first.
In this circuit, the 2.0-ohm resistor is connected in series with the three-branch parallel network. We can find the total resistance of the parallel network by using the formula:
1/R_total = 1/R1 + 1/R2 + 1/R3
In this case, R1 = R2 = R3 = 8.0 ohms. So, substituting the values into the formula:
1/R_total = 1/8.0 + 1/8.0 + 1/8.0
1/R_total = 3/8.0
Now, we can find R_total by taking the reciprocal of both sides:
R_total = 8.0/3
Now that we have the total resistance, we can find the current in the circuit using Ohm's Law. Ohm's Law states that the current (I) in a circuit is equal to the voltage (V) divided by the resistance (R):
I = V / R
In this case, the voltage of the battery is 20.0V and the total resistance (R_total) is 8.0/3 ohms. Substituting these values into the equation:
I = 20.0 / (8.0/3)
I = 20.0 * (3/8.0)
I = 7.5 Amperes
Therefore, the current in the battery is 7.5 Amperes.