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yawa3891 [41]
3 years ago
6

given a circuit powered at 12V with R1, R2, R3 respectively of 10,20,30 Ohm, determine R4 in such a way that the Wheatstone brid

ge is in equilibrium and then calculate voltages and currents as for reference exercise.​
Physics
1 answer:
elena55 [62]3 years ago
5 0

Answer:

The balanced condition for Wheatstones bridge is  

Q

P

​  

=  

S

R

​  

 

as is obvious from the given values.

No, current flows through galvanometer is zero.

Now, P and R are in series, so

Resistance,R  

1

​  

=P+R

=10+15=25Ω

Similarly, Q and S are in series, so

Resistance R  

2

​  

=R+S

=20+30=50Ω

Net resistance of the network as R  

1

​  

 and R  

2

​  

 are in parallel

i=  

R

V

​  

=  

50

6×3

​  

=0.36 A.

Explanation:

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Answer:

a) V = 252 cm³

b) Vs = 72 cm³

Explanation:

a)

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<u></u>

b)

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<u>Vs = 72 cm³</u>

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3 0
3 years ago
Read 2 more answers
If two micro coulomb of charge is flowing in a circuit for 5 minutes. What is the amount of current in the circuit?
Murrr4er [49]

Answer:

6.67×10¯⁹ A

Explanation:

From the question given above, the following data were obtained:

Quantity of electricity (Q) = 2 μC

Time (t) = 5 mins

Current (I) =?

Next, we shall convert 2 μC to C. This can be obtained as follow:

1 μC = 1×10¯⁶ C

Therefore,

2 μC = 2 μC × 1×10¯⁶ C / 1 μC

2 μC = 2×10¯⁶ C

Next, we shall convert 5 mins to seconds. This can be obtained as follow:

1 min = 60 secs

Therefore,

5 min = 5 min × 60 sec / 1 min

5 mins = 300 s

Finally, we shall determine the current in the circuit. This can be obtained as follow:

Quantity of electricity (Q) = 2×10¯⁶ C

Time (t) = 300 s

Current (I) =?

Q = It

2×10¯⁶ = I × 300

Divide both side by 300

I = 2×10¯⁶ / 300

I = 6.67×10¯⁹ A

Thus, the current in the circuit is 6.67×10¯⁹ A

4 0
2 years ago
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