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luda_lava [24]
2 years ago
10

On a box are the words 12 Volt : 50 mA What is the resistor in Kilo-Ohm?

Physics
1 answer:
katrin [286]2 years ago
6 0

Answer:

R = 0.24 kilo-ohm

Explanation:

Given that,

Voltage, V = 12 volt

Current, I = 50 mA = 0.05 A

We need to find the resistance of the resistor. We can find it using Ohm's law. So,

V = IR

R=\dfrac{V}{I}\\\\R=\dfrac{12}{0.05}\\\\R=240\ \Omega

or

R = 0.24 kohm

So, the value of resistance is 0.24 kilo-ohm.

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Current and voltage on each resistor:

I_1 = 3.98 A, V_1 = 3.98 V

I_2=0.015 A, V_2 = 0.075 V

I_3 = 0.4 A, V_3 = 0.4 V

I_4 = 0.385 A, V_4 = 0.77 V

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I_6 = 3.01 A, V_6 = 6.02 V

I_7 = 0.97 A, V_7 = 4.85 V

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega

This resistor is in series with resistor 4, so:

R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega

This resistor is in parallel with resistor 5, therefore:

R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega

This resistor is in series with resistor 7, so:

R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega

This resistor is in parallel with resistor 6, so:

R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega

Finally, this combination is in series with resistor 1:

R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A

And we can also find the potential difference across resistor 1:

V_1=I_1 R_1=(3.98)(1.0)=3.98 V

This means that the voltage across resistor 6 is

V_6=V-V_1=10-3.98=6.02 V

And so, the current on resistor 6 is

I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

I_7=I-I_6=3.98-3.01=0.97 A

And so the voltage across resistor 7 is

V_7=I_7 R_7=(0.97)(5.0)=4.85 V

The voltage across resistor 5 is

V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V

And so the current is

I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A

The current through resistor 4 is

I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A

And therefore its voltage is

V_4=I_4 R_4 = (0.385)(2.0)=0.77 V

So, the voltage through resistor 3 is

V_3=V_5-V_4=1.17-0.77=0.4 V

And the current is

I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A

Finally, the current through resistor 2 is

I_2=I_4-I_3=0.5-0.385=0.015 A

And so its voltage is

V_2=I_2R_2=(0.015)(5.0)=0.075 V

Learn more about current and voltage:

brainly.com/question/4438943

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