Question

Find the equivalent resistance, current, and voltage across each resistor when the specified resistors are connected across a 20 V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In Part (b), three resistors are specified. (a) Two resistors are connected in series across a 20 V battery, as illustrated above. Let R1

Answer 1

Answer:

Explanation:

The question is incomplete. Here is the complete question.

"Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified resistors are connected across a 20-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified. (a) Two resistors are connected in series across a 20-V battery. Let R1 = 1 Ω and R2 = 2 Ω. Rea = (b) Add a third resistor to the circuit in series. Let R1 = 1 Ω, R2 = 2 Ω, and R3 = 3 Ω"

Using ohms law formula to solve the problem

E = IRt

E is the supply voltage

I is the total current

Rt is the total equivalent resistant.

a) Given two resistances

R1 = 1ohms and R2 = 2ohms

If the resistors are Connected in series across a 20V supply voltage,

-Equivalent resistance = R1+R2

= 1ohms + 2ohms

= 3ohms

- In a series connected circuit, same current flows through the resistors.

Using the formula E = IRt

I = E/Rt

I = 20/3

I = 6.67A

The current in both resistors is 6.67A

- Different voltage flows across a series connected circuit.

Using the formula V = IR

V is the voltage across each resistor

I is the current in each resistor

For 1ohms resistor,

V = 6.67×1

V = 6.67Volts

For 2ohms resistor

V = 6.67×2

V = 13.34Volts

b) If the resistors are three

R1 = 1ohms, R2 = 2ohms R3 = 3ohms

- Total equivalent resistance = 1+2+3

= 6ohms

- Current in each resistor I = E/Rt

I = 20/6

I = 3.33A

Since the same current flows through the resistors, the current across each of them is 3.33A

- Voltage across them is calculated as shown:

V = IR

For 1ohm resistor

V = 3.33×1

V = 3.33volts

For 2ohms resistor

V = 3.33×2

V = 6.66volts

For 3ohms resistor

V = 3.33×3

V = 9.99volts

Answer 2

Complete Question

The complete question is shown on the first uploaded image

Answer:

For a

$R_{eq}= 3 \Omega$

$I = 6.667 A$

$\Delta V_1 = 6.667 \ V$

$\Delta V_2 = 13.334 \ Volt$

For  B

$R_{eq}= 6 \Omega$

$I = 3.333 A$

$\Delta V_1 = 3.333 \ Volt$

$\Delta V_2 = 6.667 \ Volt$  

$\Delta V_3 = 10 \ Volt$  

Explanation:

 From the question we are told that

           The voltage of the battery is  $V = 20 \ V$

           The  first resistance is $R_1 = 1 \Omega$

           The second resistance is $R_2 = 2 \Omega$

The equivalent resistance is mathematically represented as

           $R_{eq} = 1+2$

                  $R_{eq}= 3 \Omega$

The current  is mathematically represented as      

           $I = \frac{V}{R_{eq}}$

So

           $I = \frac{20}{3}$

            $I = 6.667 A$

The first voltage change is mathematically represented as

               $\Delta V_1 = V_1 - V_o$

               $\Delta V_1 = (R_1 * I) - 0$

               $\Delta V_1 = 6.667 \ V$

The second voltage change is mathematically represented  as

               $\Delta V_2 = V_2 - V_o$

               $\Delta V_2 = (R_2 * I ) - 0$

               $\Delta V_2 = 2* 6.667 \ Volt$

                 $\Delta V_2 = 13.334 \ Volt$    

For  B

      V = 20 V

      The  first resistance is $R_1 = 1 \Omega$  

     The second resistance is $R_2 = 2 \Omega$

     The third  resistance is $R_3 = 3 \Omega$

The equivalent resistance is  

        $R_{eq} = 1+ 2 +3$

               $R_{eq}= 6 \Omega$

The current is mathematically  evaluated as

          $I = \frac{20}{6}$

              $I = 3.333 A$

The first voltage change is mathematically represented as

               $\Delta V_1 = V_1 - V_o$

               $\Delta V_1 = (R_1 * I) - 0$

               $\Delta V_1 = (1 *3.333) \ V$

               $\Delta V_1 = 3.333 \ Volt$

The second voltage change is mathematically represented  as

               $\Delta V_2 = V_2 - V_o$

               $\Delta V_2 = (R_2 * I ) - 0$

               $\Delta V_2 = 2* 3.333 \ Volt$

                 $\Delta V_2 = 6.667 \ Volt$    

The second voltage change is mathematically represented  as

               $\Delta V_3 = V_3 - V_o$

               $\Delta V_3 = (R_3 * V ) - 0$

               $\Delta V_3 = 3 * 3.333 \ Volt$

                 $\Delta V_3 = 10 \ Volt$