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ad-work [718]
2 years ago
6

The following currents are measured in the same direction in a three-branch parallel circuit: 250 mA, 300 mA, and 800 mA. What i

s the value of the current into the junction of these three branches?
Physics
1 answer:
Ksju [112]2 years ago
5 0

The current in the junction is 1350 mA or 1.350 A.

Explanation:

As per Kirchoff's first law, the algebraic sum of current meeting at any junction should be equal to the algebraic sum of current leaving the junction. As in the present case, three parallel branch circuit is given the current in 250 mA, 300 mA and 800 mA, respectively, the sum of these three current will be equal to the current in the junction.

So,

I₁+I₂+I₃ = I₄

So I₁,I₂ and I₃ are the current passed in the three parallel branches and I₄ is the current in the junction.

250 + 300 + 800 = 1350 mA

So the current in the junction is 1350 mA or 1.350 A.

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8 0
3 years ago
An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th
neonofarm [45]

Answer:

Electric force, F=2.24\times 10^{-14}\ N

Explanation:

It is given that,

Charge on an electron is -1.6\times 10^{-19}\ C

Electric field, E=1.4\times 10^5\ N/m

We need to find the magnitude of the electric force on this electron due to this field. The electric force is given by :

F=qE\\\\F=1.6\times 10^{-19}\times 1.4\times 10^5\\\\F=2.24\times 10^{-14}\ N

So, the electric force is 2.24\times 10^{-14}\ N.

6 0
2 years ago
In the circuit shown, the galvanometer shows zero current. The value of resistance R is :
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Answer:

its supposed to be (a) 1W

8 0
3 years ago
In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot
Fittoniya [83]

Answer:

Q=7.9\times 10^{-10}\ C

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

A=5\times 10^{-3}\ m^2

We Capacitance given as

For air

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}

C_1=2.2\times 10^{-11}\ F

C_2=\dfrac{K\varepsilon _oA}{d}

C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F

C_2=6.6\times 10^{-11}\ F

Net capacitance

C=C₁+C₂

C=8.8\times 10^{-11}\ F

We know that charge Q given as

Q= C V

Q=12\times 6.6\times 10^{-11}\ C

Q=7.9\times 10^{-10}\ C

6 0
3 years ago
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