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dezoksy [38]
3 years ago
5

How to do this question?​

Physics
2 answers:
marissa [1.9K]3 years ago
7 0

Answer:

i3 =11.014A

i5 = 3.15A

Explanation:

Here according to k'chofs first law

i1 =i2 + i3

i3 = i4 + i5

For determine the i1 you have to consider the resultant resistor of the system

4 , 1 and 3 resistors are in pararel

Then, Resultant is

1/4 + 1/1 + 1/3 = 1/ R

R = 12/19

For get total we have to add another remaining 3 resistor because of serious

Then Resultant is = 12/19 + 3

= 69/19

Then using V = IR

40 =i3* 69/19

i3 = 11.014 A

Other 3 resistors are parrarel because of this voltage of those resistors are same.

Then i inversely propotional to its resistor

Then ,

i5 * 2 = (i3-i5)*4/5

i 5 = 3.15 A

Diano4ka-milaya [45]3 years ago
6 0

Answer:

I₁ = 11.2 A

I₂ = 1.6 A

I₃ = 9.6 A

I₄ = 6.4 A

I₅ = 3.2 A

Explanation:

Find the equivalent resistances.

Resistors 4 and 5 are in parallel:

R₄₅ = 1 / (1/R₄ + 1/R₅)

R₄₅ = 1 / (1/1 + 1/2)

R₄₅ = 2/3

Resistor 2 is in parallel with that:

R₂₄₅ = 1 / (1/R₂ + 1/R₄₅)

R₂₄₅ = 1 / (1/4 + 3/2)

R₂₄₅ = 4/7

Resistor 1 is in series with that:

R = R₁ + R₂₄₅

R = 3 + 4/7

R = 3 4/7

Now use Ohm's law to find I₁.

V = I₁ R

40 V = I₁ (3 4/7)

I₁ = 11.2 A

Find the voltage drop across R₁:

V₁ = I₁ R₁

V₁ = (11.2 A) (3 Ω)

V₁ = 33.6 V

So the voltage after is 40 V − 33.6 V = 6.4 V.

Find I₂:

6.4 V = I₂ (4 Ω)

I₂ = 1.6 A

Find I₃:

6.4 V = I₃ (2/3 Ω)

I₃ = 9.6 A

Find I₄:

6.4 V = I₄ (1 Ω)

I₄ = 6.4 A

Find I₅:

6.4 V = I₅ (2 Ω)

I₅ = 3.2 A

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-BARSIC- [3]

Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

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In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

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<h3>a)</h3>

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Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

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<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

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The goal is to solve for t in terms of \tau. Rearrange the equation:

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Take the natural logarithm of both sides:

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