Question

PLEASE HELP ASAP.

Answer 1

1) $32 \Omega$

2) 3.44 A

3) 27.5 V

4) $2.35\Omega$

5) 46.8 A

6) 13.7 A

Explanation:

1)

Resistors are said to be connected in series if they are connected along the same branch. In this case, the total resistance of the circuit is equal to the sum of the individual resistances:

$R=R_1+R_2+...+R_n$

In this case, we have three resistances connected in series, so the total resistance of the circuit will be:

$R=R_1+R_2+R_3$

where:

$R_1=20\Omega$ is the resistance of the 1st device

$R_2=8 \Omega$ is the resistance of the 2nd device

$R_3=4\Omega$ is the resistance of the 3rd device

Therefore, the total resistance is:

$R=20+8+4=32 \Omega$

2)

Here we want to find the current in the circuit.

First of all, we are told that the resistance of the generator is neglected: this means that the total resistance of the circuit is that found in part 1).

Therefore, we can find the current using Ohm's Law:

$V=RI$

where

V is the terminal voltage of the battery

R is the total resistance of the circuit

I is the current in the circuit

In this problem, we have:

V = 110 V is the voltage of the battery

$R=32\Omega$ is the total resistance of the  circuit

So the current is

$I=\frac{V}{R}=\frac{110}{32}=3.44 A$

3)

The voltage drop across a certain resistor is given again by Ohm's Law:

$V_i = R_i I_i$

where

$V_i$ is the voltage drop across the resistor

$R_i$ is the resistance of the resistor

$I_i$ is the current through the resistor

In this problem:

$R_2=8 \Omega$ is the resistor we are considering

$I_2=3.44 A$ is the current through the resistor (in a series circuit, the current is the same for all devices in the circuit)

So, the voltage drop is

$V_2=I_2 R_2 =(3.44)(8)=27.5 V$

4)

Electrical devices are said to be connected in parallel if they are connected along different branches having same terminal points, so that they have same potential difference across them.

The total resistance for resistors in parallel is given by:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+....+\frac{1}{R_n}$

In this problem we have 3 resistors, so the total resistance is:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

where

$R_1=20\Omega$ is the resistance of the 1st device

$R_2=8 \Omega$ is the resistance of the 2nd device

$R_3=4\Omega$ is the resistance of the 3rd device

Therefore, the total resistance is:

$\frac{1}{R}=\frac{1}{20}+\frac{1}{8}+\frac{1}{4}=0.425\\R=\frac{1}{0.425}=2.35\Omega$

5)

To find the total current in the circuit, we can use again Ohm's Law:

$V=RI$

where:

V is the terminal voltage of the battery

R is the total resistance of the circuit

I is the current in the circuit

In this problem, we have:

V = 110 V is the voltage of the battery

$R=2.35\Omega$ is the total resistance of the circuit

Therefore, solving for I, we find the current:

$I=\frac{V}{R}=\frac{110}{2.35}=46.8 A$

6)

The current across a certain resistor can be found using Ohm's Law, as before:

$V_i = R_i I_i$

where:

$V_i$ is the voltage drop across the resistor

$R_i$ is the resistance of the resistor

$I_i$ is the current through the resistor

In this problem:

$R_2=8 \Omega$ is the resistor we are considering

$V_2=110 V$ is the voltage drop across that resistor (in a parallel circuit, the voltage drop is the same across each resistor)

Solving for I, we find the current through it:

$I_2=\frac{V_2}{R_2}=\frac{110}{8}=13.7 A$