The voltage in a battery is 9 volts. The resistance is 3 ohms. What is the current?

Answer 1

There are a couple of ways to work this one. You can find the so-called branch currents (the current through each resistor) and then add them to get total current, or you can find the total effective resistance and then apply that information to the applied voltage to compute total current. Let's do this. I = E/R for each resistor. IR1 = E/6 = 9/6 = 3/2 = 1 1/2 amps IR2 = E/3 = 9/3 = 3 amps Itotal = IR1 + IR2 = (1 1/2) + (3) = 4 1/2 amps That's one way. Here's the other. Rtotal = 1 / [(1/R1) + (1/R2)] = 1 / [(1/6) + (1/3)] = 1 / [(1/6) + (2/6)] = 1 / [3/6] = 2 ohms Itotal = Eapplied / Rtotal = 9 volts / 2 ohms = 4 1/2 amps It checks. What a surprise. Ohm's law is correct.

Answer 2

It depends upon whether 3 ohms is its 'cold' or 'hot' resistance. The resistance of an incandescent lamp filament can increase up to x15 as its temperature rises from cold to its operational value.

Answer 3

Two 6-ohm resistors in parallel have a net effective resistance of 3 ohms.

With 3 ohms connected across a 12-volt supply, the current is 12/3 = 4 amperes.

Answer 4

Ohm's Law: Current is voltage divided by resistance

9 volts divided by 3 ohms is 3 amperes.

Answer 5

Though you have connected 6 ohm parallel to 12 V cell, it is only a series circuit.

Now current = volt / resistance = 12/6 = 2 ampere.

Answer 6

The current flow through the circuit is , 9 Volts/3 Ohms= 3 Amperes.