Question

1180 resistance and60reactance are connected in
series in an AC circuit. Then the impedance of the
circuit is
यदि प्रतिरोचतथा 60 प्रतिघातको एक AC परिपथ में श्रेणी में
संयोजित किया जाता है। तब परिपथ की प्रतिबाधा है
140
100
20
10202​

Answer 1

Answer:

1178.4 $\Omega$

Explanation:

Reactance = $\\\tt \sqrt{(reistance)^2 + (reactance)^2 }$

Using the given values:

$\\\\\tt impedance \ = \sqrt{1180^2 + 60^2} \\\\\tt = \sqrt{1388800} \\\tt = 1178.4 \Omega$

Answer 2

Given :-

1180  resistance and 60 reactance are connected in  series in an AC circuit

To Find :-

Impedance

Solution :-

$\sf Impedance = \sqrt{(r)^2 + (r)^2}$

By putting value

$\sf Impedance = \sqrt{(1180)^2 + (60)^2}$

$\sf Impedance= \sqrt{139200+3600}$

$\sf Impedance = \sqrt{1388800}$

$\sf Impedance= 1178.4 \; \Omega$