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3 years ago

For a constant resistance, how is the voltage related to the current?

2 answers:

6 0

The answer is: <span> Voltage is directly proportional to current, so when the voltage doubles, the current doubles (V=I*R).</span>

The relationship between voltage, current, and resistance is presented by Ohm's law.

STALIN [3.7K]3 years ago

5 0

Answer:
Voltage is directly proportional to current, so when the voltage doubles, the current doubles.
Explanation:
voltage, current and resistance can be related to each other using Ohm's law which can be stated as follows:
V = IR where:
V is teh voltage
I is teh current
R is the resistance

Now, assuming that the value of R is constant, we can note that as the current increases, the voltage increases by the same value of increase and vice-versa

This means that, at constant resistance, the voltage and current are directly proportional to each other.

Hope this helps :)

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Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th

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Explanation:

As the given data is as follows.

$R_{1} = 680 \ohm$ ohm $\ohm$ , $R_{2} = 720 \ohm$ ohm,

$R_{3} = 1.2 k\ohm$ = 1200 $\ohm$ (as 1 k ohm = 1000 m)

(a) We will calculate the maximum resistance by combining the given resistances as follows.

Max. Resistance = $R_{1} + R_{2} + R_{3}$

= $(680 + 720 + 1200) \ohm$ ohm

= 2600 ohm

or, = 2.6 $k\ohm$ ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 $k\ohm$ ohm.

(b) Now, the minimum resistance is calculated as follows.

Min. Resistance = $\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

= $\frac{1}{680} + \frac{1}{720} + \frac{1}{1200}$

= $3.683 \times 10^{-3}$ ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is $3.683 \times 10^{-3}$ ohm.

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