High School

We appreciate your visit to What is the current in a 3 632 Omega resistor connected to a battery that has a 0 1382 Omega internal resistance when the potential. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

What is the current in a \( 3.632 \, \Omega \) resistor connected to a battery that has a \( 0.1382 \, \Omega \) internal resistance when the potential drop across the terminals of the battery is \( 6 \, V \)?

Answer :

The current in a 3.632 ohm resistor connected to a battery with 0.1382 ohms internal resistance and a 6V potential drop across its terminals is approximately 1.592 A.

To find the current in a resistor connected to a battery, we can use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the total resistance (Rtotal). The total resistance is the sum of the external resistance and the battery's internal resistance. In this case, the potential drop across the terminals of the battery is given as 6V, the external resistance is 3.632 ohms, and the internal resistance of the battery is 0.1382 ohms.

First, we calculate the total resistance:

Rtotal = external resistance + internal resistance

= 3.632 ohms + 0.1382 ohms

= 3.7702 ohms.

Next, we use Ohm's Law to find the current:

I = V / Rtotal

= 6V / 3.7702 ohms.

Therefore, the current in the circuit is approximately 1.592 A.

Thanks for taking the time to read What is the current in a 3 632 Omega resistor connected to a battery that has a 0 1382 Omega internal resistance when the potential. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada