Question

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. Perform calculations using the circuit illustrated. Round all the numerical answers to the tenths place. The total resistance in the circuit is Ω. The expected current at point A is A. At what point in the circuit will the current be the lowest (A, B, C, or D)?

Answer 1

Answer:

Current: 1.0 Amperes

The minimum current is flowing through path D

Explanation:

We first find the equivalent resistance to the three resistors in parallel ( which is the total resistance of the circuit) via the equation:

$\frac{1}{R_e} =\frac{1}{R_B}+\frac{1}{R_C}+\frac{1}{R_D}\\\frac{1}{R_e} =\frac{1}{10}+\frac{1}{20}+\frac{1}{50}=0.17\\R_e=(1/0.17)\Omega\\R_e=5.88 \Omega$

with this info, we can estimate the current going through branch A using Ohm's Law, and the information that the power source is 6 V:

$V=I*R\\6V=I*5.88\Omega\\I=\frac{6}{5.88} Amp\\I=1.02A$

where the current comes in units of Amperes since all other the quantities are given in the SI system, and we can round this answer to 1.0 Amp following the request to round it to the tenth.

The current will be the lowest through the branch with the largest resistor due to the fact that less current will flow through the path of more resistance.

Than means that the lowest current will be registered through branch D where the 50 $\Omega$ resistor is.

Answer 2

Answer:

1.) 59

2.)1

3.)D

Explanation:

right on e2020