Question

Three resistors of equal resistance are connected in series across a voltage of 9 V. If the current through the resistors is measured as 1 A, what is the resistance of each resistor? If the middle resistor is replaced with a conductor of negligible resistance, what is the current through the circuit? Each resistor is Ω. If the middle resistor is replaced, the current would be A

Answer 1

If three equal resistors are connected in series and the total voltage across them is 9 V, and the current through the circuit is 1 A, we can use Ohm's law to calculate the resistance of each resistor.

Ohm's law states that V (voltage) = I (current) * R (resistance).

So, the resistance of each resistor (R) can be calculated as:

R = V / I = 9 V / 1 A = 9 Ω.

So, each resistor has a resistance of 9 ohms.

Now, if the middle resistor is replaced with a conductor of negligible resistance, the total resistance in the circuit becomes:

Total Resistance = 9 Ω (first resistor) + 0 Ω (conductor) + 9 Ω (third resistor) = 18 Ω.

Now, to calculate the current (I) in the circuit with the conductor, we can use Ohm's law again:

I = V / R = 9 V / 18 Ω = 0.5 A.

So, if the middle resistor is replaced with a conductor, the current through the circuit would be 0.5 A.