refer to the following for questions 4 - 5: 4. if the current at point 1 in the series circuit above is 3 a…

refer to the following for questions 4 - 5: 4. if the current at point 1 in the series circuit above is 3 a and the voltage drop across the first, second, and third resistors is 2 v, 3 v, and 4 v, respectively, what is the total resistance of the circuit? a. 6 ω b. 3 ω c. 9 ω d. 0.33 ω 5. if the current at point three in the series circuit above is 4 a, the resistance of r₁ is 0.5 ω, the resistance of r₂ is 1 ω, and the voltage drop across r₃ is 6 v, what is the size of the battery of the circuit? a. 12 v b. 1.5 v c. 9 v d. 3 v refer to the following for questions 6 - 8: 6. in the above circuit, what is the current at points 4 and 6 if the battery is 20 v, r₂ = 4 ω, and r₃ = 2 ω? a. i₄ = 10 a; i₆ = 15 a b. i₄ = 20 a; i₆ = 20 a c. i₄ = 10 a; i₆ = 10 a d. i₄ = 15 a; i₆ = 5 a 7. in the above circuit, what is the equivalent resistance if r₁ = 4 ω, r₂ = 4 ω, and r₃ = 2 ω? a. 10 ω b. 3.33 ω c. 1 ω d. 32 ω

refer to the following for questions 4 - 5: 4. if the current at point 1 in the series circuit above is 3 a and the voltage drop across the first, second, and third resistors is 2 v, 3 v, and 4 v, respectively, what is the total resistance of the circuit? a. 6 ω b. 3 ω c. 9 ω d. 0.33 ω 5. if the current at point three in the series circuit above is 4 a, the resistance of r₁ is 0.5 ω, the resistance of r₂ is 1 ω, and the voltage drop across r₃ is 6 v, what is the size of the battery of the circuit? a. 12 v b. 1.5 v c. 9 v d. 3 v refer to the following for questions 6 - 8: 6. in the above circuit, what is the current at points 4 and 6 if the battery is 20 v, r₂ = 4 ω, and r₃ = 2 ω? a. i₄ = 10 a; i₆ = 15 a b. i₄ = 20 a; i₆ = 20 a c. i₄ = 10 a; i₆ = 10 a d. i₄ = 15 a; i₆ = 5 a 7. in the above circuit, what is the equivalent resistance if r₁ = 4 ω, r₂ = 4 ω, and r₃ = 2 ω? a. 10 ω b. 3.33 ω c. 1 ω d. 32 ω

Answer

4.

Explanation:

Step1: Find total voltage

In a series - circuit, the total voltage $V_{total}$ is the sum of the voltage drops across each resistor. So $V_{total}=2 + 3+4=9$ V.

Step2: Use Ohm's law

Ohm's law is $V = IR$, where $V$ is voltage, $I$ is current, and $R$ is resistance. We know $I = 3$ A and $V=9$ V. Rearranging for $R$, we get $R=\frac{V}{I}$. $R=\frac{9}{3}=3$ $\Omega$.

Answer:

b. $3\Omega$

5.

Explanation:

Step1: Find voltage drops across $R_1$ and $R_2$

Using Ohm's law $V = IR$. For $R_1$ with $I = 4$ A and $R_1=0.5$ $\Omega$, $V_1=4\times0.5 = 2$ V. For $R_2$ with $I = 4$ A and $R_2 = 1$ $\Omega$, $V_2=4\times1=4$ V.

Step2: Find total voltage

The total voltage of the battery $V_{total}$ in a series - circuit is the sum of the voltage drops across all resistors. So $V_{total}=V_1 + V_2+V_3$, where $V_3 = 6$ V. Then $V_{total}=2 + 4+6=12$ V.

Answer:

a. $12$ V

6.

Explanation:

Step1: Analyze parallel - circuit properties

In a parallel circuit, the voltage across each branch is equal to the battery voltage. Here $V = 20$ V. For $R_2 = 4$ $\Omega$ and $R_3=2$ $\Omega$, using Ohm's law $I=\frac{V}{R}$. The current through $R_2$, $I_2=\frac{20}{4}=5$ A. The current through $R_3$, $I_3=\frac{20}{2}=10$ A. In a parallel circuit, the current at point 4 is the current through $R_3$ which is $I_4 = 10$ A, and the current at point 6 is the current through $R_2$ which is $I_6 = 10$ A (since the voltage across each parallel - branch is the same and we are considering the currents in the respective branches).

Answer:

c. $I_4 = 10$ A; $I_6 = 10$ A

7.

Explanation:

Step1: Use formula for parallel resistors

The formula for the equivalent resistance $R_{eq}$ of three parallel resistors $R_1$, $R_2$, and $R_3$ is $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$. Given $R_1 = 4$ $\Omega$, $R_2 = 4$ $\Omega$, and $R_3=2$ $\Omega$. $\frac{1}{R_{eq}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=\frac{1 + 1+2}{4}=\frac{4}{4}=1$. Then $R_{eq}=1$ $\Omega$.

Answer:

c. $1$ $\Omega$