Question

two reisitors (a and b) are connected in parallel to a 12-v battery. resistor a has 2.0 a in it and the total current in the battery is 3.0 a. which resisitor has the most resistance?

Answer 1

Resistor "b".

Step-by-step explanation:

Since the two resistors are connected in parallel:

• Voltage across the two resistors would be the same, and
• Sum of the current in the two resistors would be equal to the total current in the circuit.

Since both resistors are connected directly to the
`12\; {\rm V}` battery, the voltage across both resistors would be
`V = 12\; {\rm V}`.

Current in resistor "a" is
`2.0\; {\rm A}` while the total current is
`3.0\; {\rm A}`. Hence, current in the other resistor (resistor "b") would be
`3.0\; {\rm A} - 2.0\; {\rm A} = 1.0\; {\rm A}`.

Apply Ohm's Law to find the resistance of each resistor. By Ohm's Law:

`\displaystyle R = (V)/(I)`,

Where:

• 
  `V` is the voltage across the resistor, and
• 
  `I` is the current in the resistor.

The resistance of resistor "a" would be:

`\begin{aligned}R &= \frac{12\; {\rm V}}{2.0\; {\rm A}} = 6.0\; {\rm \Omega} \end{aligned}`.

The resistance of resistor "b" would be:

`\begin{aligned}R &= \frac{12\; {\rm V}}{1.0\; {\rm A}} = 12\; {\rm \Omega} \end{aligned}`.

Hence, resistance of resistor "b" is higher than that of resistor "a".