All questions of Network Basics & Basic Laws for Electrical Engineering (EE) Exam

Total current = 150/(6+12+15) = 4.55A.

V across 6 ohm = Total current x resistance

= 4.55 * 6 = 27.24V.

The 1 ohm and 2-ohm resistor are in series which is in parallel to the 3-ohm resistor.
The equivalent of these resistances is in series with the 4 ohms and 5-ohm resistor.
Total R= 10.5 ohm. I=V/R= 120/10.5= 11.42A.

Explanation : Current due to voltage source is:

I' = 10/(64+36) 

= 0.1A.

Current due to current source is:

I = 0.5A.

Current ib = I' + I 

= 0.5+ 0.1

= 0.6A

I = V/R.

Total resistance = 20+40 = 60ohm.

I = 120V.

I = 120/60 = 2A.

The 60ohm resistance is shorted since current always choses the low resistance path. Voltage across short circuit is equal to zero, hence voltage across the resistor is 0.

The 5 ohm and 15 ohm resistances are connected in series and 10 ohm and 20 ohm resistors are also connected in series. Therefore,20 ohm and 30 ohm  are parallel with each other.

Going from 10 V to 0 V

10 + 5 + E + 1 + 0 = 0 or E =-16 V

The two mesh equations are:
5I1-3I2=10
-3I1+7I2=-15
Solving the equations simultaneously, we get I1=0.96A and I2=-1.73A.

Total voltage=sum of voltages across each resistor. =>150=10*5+5*5+5*x. Solving the equation, we get x=15 ohm.

The resistors are connected in parallel, hence the equivalent resistance= 1/(1/20=1/20+1/20+1/20)=5ohm.

 R=((2+3)||5)+1.5)||4.The 2 and the 3 ohm resistor are in series. The equiva-lent of these two resistors is in parallel with the 5 ohm resistor. The equivalent of these three resistances is in series with the 1.5 ohm resistor. Finally, the equivalent of these re-sistances is in parallel with the 4 ohm resistor.

Mesh analysis helps us to utilize the different voltages in the circuit as well as the IR products in the circuit which is nothing but KVL.

You can know more about Kirchhoff's Laws  through the document: 

To explain: Using KVL in loop 1, 10-100 i1=0. i1=0.1A

Using KVL in outer loop, -100i2+20=0  i2=0.2A.

Correct Answer is d)  0.1, 0.2

Kirchhoff’s Current Law

Kirchhoff’s Current Law (KCL) is Kirchhoff’s first law that deals with the conservation of charge entering and leaving a junction. To determine the amount or magnitude of the electrical current flowing around an electrical or electronic circuit, we need to use certain laws or rules that allows us to write down these currents in the form of an equation. The network equations used are those according to Kirchhoff’s laws, and as we are dealing with circuit currents, we will be looking at Kirchhoff’s current law, (KCL).

 In a parallel circuit, if one bulb blows out, it acts as an open circuit. Current does not flow in that branch but it continues to flow in the other branch hence the bulb continues to glow.

Explanation: I=V/R. Since in parallel circuit, voltage is same across all resistors. Hence across the 20 ohm resistor, V=200V so I=200/20=10A.

The voltage across a short is always equal to zero whether it is connected in seroes or parallel.

KCl states that the total current leaving the junction is equal to the current entering it. In this case, the current entering the junction is 5A+10A=15A.

Batteries are generally connected in series so that we can obtain the desired voltage since voltages add up once they are connected in series.

According to Kirchhoff’s Voltage Law, Every mesh is a loop but every loop is not a mesh.

The current through the 10 ohm resistor = v₁ / 10 = 2A

Applying KCL at node 1: i₅ = i₁₀ + i₂

i₂ = 6 - 2 = 4A

Thus the drop in the 2 ohm resistor = 4×2 = 8V

v₁ = 20V; hence v₂ = 20-v across 2 ohm resistor = 20 - 8 = 12V
v₂ = v since they are connected in parallel.
v = 12V

If we go from +side of 1 kΩ through 7 V, 6 V and 5V, we get v₁ = 0 +  7 + 6 - 5 = 8 V

Voltage is constant because of 15 V source.

Using KVL, 12-V1-8=0. V1= 4V.
8-V2-2=0. V2=6V.

Since the two bulbs are connected in series, if the first bulb burns out there is a break in the circuit and hence the second bulb does not glow.

I=V/R. Hence if R=2R V=I/2R, and I=I/2.

Let's calculate the charge on the sphere

Understanding the problem:

Solution:

Total charge (Q) = n * e

Q = 1020 * (-1.6 x 10^-19 C) Q = -1.632 x 10^-16 C

Approximately, Q = -16.02 x 10^-17 C

So, the correct answer is:

Note that the charge is negative because electrons are removed, leaving a net positive charge on the sphere.

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Using voltage divider rule, V=10*12/30=4V.

 In parallel circuits, the current across the circuits vary whereas the voltage remains the same.

Reraw the circuit:

Now, using voltage division,

The node equation is:
-2+8+V/10=0 => 6 + v/10 = 0 => v = 10*6=>60v
Solving this equation, we get V=60V.

A short is just a wire. The potential difference between two points of a wire is zero hence the voltage measured is equal to zero.

W = Qv = 360J

KCL states that the amount of charge leaving a node is equal to the amount of charge entering it, hence it is applied at nodes.

i₁ = 0.02 A

Sign convention for Power delivered and absorbed:
When current is leaving the positive voltage terminal of the element or current is entering the negative voltage terminal of the element then the particular element is delivering power.
∴ P = +Vi

∴ P = -Vi

Calculation:
Given that, 
V = 2 volts, 
i = 3 A

In the above circuit, the current is leaving the positive voltage terminal of the element or the current is entering the negative voltage terminal of the element then the particular element is delivering power.
Thus, the power drawn by the circuit shown in the above figure can be calculated as
P = + Vi
=> P = +(2 × 3)
∴ P = + 6 W

In order for 600 C charge to be delivered to the 100 V source, the current must be anticlockwise.

Applying KVL we get