Computer models can be touched. True False

Solving a series circuit, did I do this correctly?

nirvana33 [79]

The total resistance in the circuit is 16 Ohms.
The total current in the circuit is 0.5 Ampere.
The current at $R_1$ is 0.5 Ampere.
The current at $R_3$ is 0.5 Ampere.
The voltage drop at $R_1$ is 4 volts.
The voltage drop at $R_2$ is 2.5 volts.
The voltage drop at $R_3$ is 1.5 volts.
The total power consumed by the circuit is 4watts
The power consumed at $R_1$ is 2 watts
The power consumed at $R_2$ is 1.25 watts

Given:

The voltage across the circuit = V = 8 V

The resistors connected are in series:

$R_1=8 \Omega, R_2=5\Omega ,R_3=3 \Omega$

To find:

The values of from 1 to 10.

Solution

The voltage across the circuit = V = 8 V

The total resistance in the circuit = $R_{eq}$

$R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega$

The total current in the circuit = I

$V=IR_{eq}\\I=\frac{V}{R_{eq}}=\frac{8 V}{16 \Omega}=0.5 A$ (Ohm's law)

For series combinations, the current in each resistor remains the same.

So, the current in $R_1, R_2 \&R_3$ :

$I_1= I_2= I_3=I=0.5 A\\$

The voltage drop across at $R_1$ = $V_1$

The current across $R_1$ = I = 0.5 A

$V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V$

The voltage drop across at $R_2$ = $V_2$

The current across $R_2$ = I = 0.5 A

$V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V$

The voltage drop across at $R_3$ = $V_3$

The current across $R_3$ = I = 0.5 A

$V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V$

The total power consumed by circuit:

$P= V\times I \\\\= 0.5 A\times 8 V = 4 watt$

Power consumed at $R_1$ :

$P_1=V_1\times I\\\\= 4V\times 0.5A = 2 watt$

Power consumed at $R_2$ :

$P_2=V_2\times I\\\\= 2.5 V\times 0.5A = 1.25 watt$

Power consumed at $R_3$ :

$P_3=V_3\times I= \\\\1.5 V\times 0.5A = 0.75 watt$

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