Using Ohm's Law, we can calculate the resistance:
Resistance = Voltage / Current
Resistance = 9 V / 3 A = 3 ohms
Therefore, the resistance in the circuit is 3 ohms.
Calculate the resistance in a circuite in which a 9-volt battery produces 3 amperes. What is the ohms?
9 answers
yes the answer is a. 3 ohms RT= 9volts/3amps=3 V=9volts, and I=3amps:R=3ohms
Thank you for confirming the answer! Yes, the resistance is indeed 3 ohms when the voltage is 9 volts and the current is 3 amps.
There was a chart, it needs the information changed by only piece : change the battery voltage to 9 volts.Once you have redone each of the tables, what is the new voltage across light bulb #2? I have 5V as an answer.
Without seeing the original tables, it is difficult to provide a definitive answer. However, if the original voltage across light bulb #2 was 3 volts and the battery voltage was changed to 9 volts, then it is likely that the voltage across light bulb #2 would also change.
Assuming the resistance of the circuit and the current flowing through it remain constant, we can use Ohm's Law to calculate the new voltage across light bulb #2:
V = IR
where V is the voltage, I is the current, and R is the resistance.
If the current through the circuit is also 3 amps, then:
V = 3A x R
If the resistance of light bulb #2 is, say, 1 ohm, then:
V = 3A x 1 ohm = 3 volts
However, if the battery voltage is increased to 9 volts, then:
V = 3A x 1 ohm = 3 volts x (9 volts / 3 volts) = 9 volts
So it is possible that the new voltage across light bulb #2 would be 9 volts instead of 5 volts. Again, without seeing the original tables, it is difficult to say for sure.
Assuming the resistance of the circuit and the current flowing through it remain constant, we can use Ohm's Law to calculate the new voltage across light bulb #2:
V = IR
where V is the voltage, I is the current, and R is the resistance.
If the current through the circuit is also 3 amps, then:
V = 3A x R
If the resistance of light bulb #2 is, say, 1 ohm, then:
V = 3A x 1 ohm = 3 volts
However, if the battery voltage is increased to 9 volts, then:
V = 3A x 1 ohm = 3 volts x (9 volts / 3 volts) = 9 volts
So it is possible that the new voltage across light bulb #2 would be 9 volts instead of 5 volts. Again, without seeing the original tables, it is difficult to say for sure.
The choice of answer is either a. 6v b.5v c.4v d.3v so The battery voltage change is 9volts and the new voltage across light bulb #2 is 3v.
Thank you for providing the answer choices. If the answer choices are between 6 volts, 5 volts, 4 volts, and 3 volts, and the only change was the battery voltage from an unknown value to 9 volts, then it is likely that the resistance and current values have remained constant as well.
In that case, we can use the voltage divider rule to calculate the voltage across light bulb #2:
V2 = (R2 / (R1 + R2)) x Vtotal
where Vtotal is the total voltage (9 volts in this case), R1 is the resistance of the first bulb, and R2 is the resistance of the second bulb.
Assuming the resistance of the first bulb is 2 ohms and the resistance of the second bulb is 1 ohm, then:
V2 = (1 ohm / (2 ohms + 1 ohm)) x 9 volts = 3 volts
Therefore, the new voltage across light bulb #2 would be 3 volts (choice d).
In that case, we can use the voltage divider rule to calculate the voltage across light bulb #2:
V2 = (R2 / (R1 + R2)) x Vtotal
where Vtotal is the total voltage (9 volts in this case), R1 is the resistance of the first bulb, and R2 is the resistance of the second bulb.
Assuming the resistance of the first bulb is 2 ohms and the resistance of the second bulb is 1 ohm, then:
V2 = (1 ohm / (2 ohms + 1 ohm)) x 9 volts = 3 volts
Therefore, the new voltage across light bulb #2 would be 3 volts (choice d).
yes the answer is d. 3v V2=(R2/(R1+R2) ) xVtotal
Great! Glad to be of help. Yes, the voltage divider rule can be a useful tool for determining the voltage across specific components in a circuit. The answer, using the values you provided, is indeed 3 volts.