If the resistance is added in parallel to a circuit, the circuit resistance is than?

Physics

2 answers:

Tcecarenko [31]3 years ago

5 0

Answer:

The total resistance in a parallel circuit is always less than any of the branch resistances. Adding more parallel resistances to the paths causes the total resistance in the circuit to decrease. As you add more and more branches to the circuit the total current will increase because Ohm's Law states that the lower the resistance, the higher the current.

Explanation:

inna [77]3 years ago

5 0

Answer:

Decreased.

Explanation:

When multiple resistors are connected in such a way that their two terminals share the same node. In such a connection the total resistance is smaller than the resistance value of any of the resistors. If the resistance of circuit is R and a resistor with resistance value R' is connected in parallel to it, the combined resistance will be,

$\frac{1}{R_{New}}=\frac{1}{R}+\frac{1}{R'}$

You might be interested in

HELP ASAP PLEASEEEEEEEEEEEE

ira [324]

It's the second graph!
it's the only one with a negative gradient.
so the temperature of the ball will fall in water as it looses its heat.

activate windows,:-P

8 0

3 years ago

Two asteroids are 75000 m apart one has a mass of 8 x 10^7 kg if the force of gravity between them is 1.14 what is the mass of t

vaieri [72.5K]

I’m going to say B

I hope this helped

4 0

3 years ago

Read 2 more answers

A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times

kodGreya [7K]

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

$L=10log_{10}\left ( \frac{I}{I_0}\right )$

A train whistle has a sound intensity level of 70 dB

We have

$70=10log_{10}\left ( \frac{I_1}{I_0}\right )$

A library has a sound intensity level of about 40 dB

We also have

$40=10log_{10}\left ( \frac{I_2}{I_0}\right )$

Dividing both equations

$\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000$