good or bad shsjajajsjs
Answers
Answer: I don't know good?? shsjsjsjsjsjsjsjs
Answer:
{\huge {\bf {\underbrace {Question}}}}
Question
An electric lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to same source that takes as much current as all three appliances and what is the current through it?
{\huge {\bf {\underbrace {Answer}}}}
Answer
Given,
Resistance of the electric lamp {\sf {{R}_{1}}}R
1
= 100 Ω
Resistance of toaster, {\sf {{R}_{2}}}R
2
= 50 Ω
Resistance of water filter, {\sf {{R}_{3}}}R
3
= 500 Ω
Since the three appliances are connected in parallel therefore their equivalent resistance is
{\large {\boxed {\pink {\sf {\frac{1}{{R} _ {p}} = \frac{1}{{R} _ {1}} + \frac{1}{{R} _ {2}} + \frac{1}{{R} _ {3}}}}}}}
R
p
1
=
R
1
1
+
R
2
1
+
R
3
1
{\large {\sf {\implies {\frac{1}{{R}_{p}} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500}}}}}⟹
R
p
1
=
100
1
+
50
1
+
500
1
{\large {\sf {\implies {\frac{1}{{R}_{p}} = \frac{5+10+1}{500}}}}}⟹
R
p
1
=
500
5+10+1
{\large {\sf {\implies {\frac{1}{{R}_{p}} = \frac{16}{500}}}}}⟹
R
p
1
=
500
16
{\green {\boxed {\sf {\implies {{R}_{p} = \frac{125}{4}Ω}}}}}
⟹R
p
=
4
125
Ω
Now,
current through the three appliances is
{\boxed {\sf {\pink {I = {\frac{V}{R}}}}}}
I=
R
V
{\large {\sf {\implies {I = 220 \times {\frac{4}{125}}}}}}⟹I=220×
125
4
{\green {\boxed {\sf {\implies {I = 7.04\: A}}}}}
⟹I=7.04A
Since, the electric iron is connected to the same source (i.e., 220 V) and takes as much current as all three appliances i.e., I = 7.04 A, therefore its resistance is equal to
{\large {\pink {\boxed {\sf {R = {\frac{V}{I}}}}}}}
R=
I
V
{\large {\sf {\implies {R = {\frac{220}{7.04}}}}}}⟹R=
7.04
220
{\sf {\green {\boxed {\implies {R = 31.25 Ω}}}}}
⟹R=31.25Ω
Current through the electric iron, I = 7.04 A.