Test: Current Electricity - 1 - CUET Humanities MCQ

The condition for no deflection of the galvanometer is:

where R_AC and R_CB are the resistances of the bridge wire of lengths AC and CB, respectively. If the radius of the wire AB is doubled, the ratio R_AC : R_CB will remain unchanged. Hence, the balance length will remain the same. Thus, the correct choice is (2).

Current I will be independent of the resistance R₆ if resistance R₆ is ineffective. This will be possible only when resistances R₁, R₂, R₃ and R₄ form a balanced Wheatstone bridge, i.e.

Electricity is the flow of charge. Usually, our charges will be carried by free-flowing electrons. Negatively-charged electrons are loosely held to atoms of conductive materials. With a little push, we can free electrons from atoms and get them to flow in a generally uniform direction.

Potential difference across PQ i.e., potential difference across the resistance of 20Ω, which is V = i × 20
∴

To get maximum effective resistance, the two resistors should be combined in series.
As effective resistance = sum of two resistances,
e.g. R = R₁ + R₂
To get minimum effective resistance, the two resistors should be combined in parallel.
e.g. R = 

Using delta to star conversion formula, we get R_AB = R ohm.
Current flowing from point A will be equal to V/R Amp.
But this total current will get split into two equal parts as the equivalent resistance of the two branches is the same.
So, current flowing through AFCEB = Half of the total current flowing from point A =(1/2)(V/R) = V/2R Amp

According to question,
E - Ir₁ = 0

2r₁ = r₁ + r₂ + R
R = r₁ - r₂

We can simplify the network as shown:

So, net resistance,
R = 2.4 + 1.6 = 4.0Ω
Therefore, current from the battery,
i = V/R = 4/4 = 1A
Now from the circuit (b),
4l' = 6I

Resistance R bisecting the circuit can be neglected due to the symmetry of the circuit. Now there are four triangles. Effective resistance of each triangle

Now the given circuit is reduced to 

Therefore, effective resistance between A and B,