A 65.0-Ù resistor is connected in parallel with a 117.0-Ù resistor. This parallel group is connected in series with a 22.0-Ù resistor. The total combination is connected across a 15.0-V battery
Find the current in the 117.0-Ù resistor.
Find the power dissipated in the 117.0-Ù resistor
total resistance parallel branch:
Req=65*117/(182)= 41.8 ohms
total circuit resistance: 41.8+22=63.8 ohms
total current: 15/63.8=.235a
voltage across the parallel branch:
.235*41.8=9.82 volts
current in 117 = 9.82/117 amps
To find the current in the 117.0-Ù resistor, we need to first calculate the total resistance of the parallel combination.
To calculate the total resistance of a parallel combination of resistors, we use the formula:
1 / Rt = 1 / R1 + 1 / R2 + 1 / R3 + ...
In this case, we have two resistors in parallel: 65.0-Ù and 117.0-Ù. So, we can calculate the total resistance of the parallel combination:
1 / Rt = 1 / 65.0 + 1 / 117.0
Now we can find Rt by taking the reciprocal of the sum:
Rt = 1 / (1 / 65.0 + 1 / 117.0)
Calculate Rt, and then we can move on to finding the current in the 117.0-Ù resistor.
To find the current in the 117.0-Ù resistor, we use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by the resistance (R) of the resistor:
I = V / R
In this case, we have the voltage (V) of 15.0 V from the battery and the resistance (R) of the 117.0-Ù resistor. Now we can calculate the current:
I = 15.0 V / 117.0-Ù
Calculate this expression, and you will find the current in the 117.0-Ù resistor.
To find the power dissipated in the 117.0-Ù resistor, we use the formula:
P = I^2 * R
In this case, we have already calculated the current (I) in the 117.0-Ù resistor, and we have the resistance (R) of the resistor. Now we can calculate the power:
P = (current)^2 * 117.0-Ù
Calculate this expression, and you will find the power dissipated in the 117.0-Ù resistor.