If a 12v battery is in a circuit with two resistors. r1 and r2. R2 is 60ohms. A voltmeter measuring on either side of r2 gives a reading of 8v.
Calculate the current in the circuit and the resistance of r1.
Am I right in thinking the v = i/r
8 = 7.5 / 60
so the current is 7.5amps
Would this mean I could work out the total resistance by multiplying the original 12v by the current?
12 x 7.5 = 90 meaning r1 = 30?
NO. V=IR, so I=V/R=8/60 amps
To calculate the current in the circuit, you use Ohm's Law, which states that V = IR, where V is the voltage, I is the current, and R is the resistance. In this case, the voltage across resistor R2 is given as 8V, and the resistance of R2 is given as 60 ohms. So, to find the current, you rearrange the formula to solve for I: I = V/R.
Therefore, I = 8V / 60 ohms = 0.1333 A (approximately).
Now, to find the resistance of R1, we need to determine the total current in the circuit. The total resistance in a series circuit is the sum of individual resistances, so we can use Ohm's Law again: V = IR. In this case, the voltage across the whole circuit is 12V, and we just calculated the current to be 0.1333A.
So, 12V = 0.1333A x (R1 + 60 ohms).
To find R1, rearrange the formula to solve for R1:
R1 = (12V - 0.1333A x 60 ohms) = 11.6V / 0.1333A = 87 ohms (approximately).
Therefore, the current in the circuit is approximately 0.1333A, and the resistance of R1 is approximately 87 ohms.
Your calculation for the current is almost correct. However, you mentioned using 7.5A instead of 0.1333A. Also, multiplying the original 12V by the current will give you the power (P = VI), not the resistance. To find the resistance, you need to rearrange Ohm's Law and use the appropriate values.