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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?

User Lanna
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1 Answer

2 votes

Answer:

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".

User Zorglub
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