A diagram of a closed circuit with a power source on the left labeled 6 V. There are 3 resistors in parallel, separate paths, connected to it labeled B 10 Ohms, C 20 Ohms and D 50 Ohms. The branch with the power source is labeled A. There are vectors in the clockwise direction across all three resistors and in the branch with the power source.

Question

Perform calculations using the circuit illustrated. Round all the numerical answers to the tenths place.

The total resistance in the circuit is

Answer 1

To calculate the total resistance in a parallel circuit, we use the formula for total resistance \( R_t \) in parallel:

\[ \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

where \( R_1, R_2, \) and \( R_3 \) are the resistances of the individual resistors. In your case, we have:

• \( R_B = 10 , \Omega \)
• \( R_C = 20 , \Omega \)
• \( R_D = 50 , \Omega \)

Now let's plug in the values into the formula:

\[ \frac{1}{R_t} = \frac{1}{10} + \frac{1}{20} + \frac{1}{50} \]

Calculating each term:

• \( \frac{1}{10} = 0.1 \)
• \( \frac{1}{20} = 0.05 \)
• \( \frac{1}{50} = 0.02 \)

Now, add these values together:

\[ \frac{1}{R_t} = 0.1 + 0.05 + 0.02 = 0.17 \]

Next, take the reciprocal to find the total resistance \( R_t \):

\[ R_t = \frac{1}{0.17} \approx 5.8824 , \Omega \]

Rounding to the tenths place:

\[ R_t \approx 5.9 , \Omega \]

Therefore, the total resistance in the circuit is approximately 5.9 Ohms.