Current Electricity

Two bulbs are rated "100 W – 220 V" and "60 W – 220 V". When connected in SERIES across a 220 V supply, which bulb glows brighter (dissipates more power)?

1. The 100 W bulb, since it has the higher power rating
2. The 60 W bulb, since it has higher resistance, and in series (same current through both), power P = I²R is larger for the larger resistance
3. Both glow with exactly equal brightness, since the power ratings simply add in series
4. Neither glows, since operating below rated voltage destroys both bulbs immediately

The same galvanometer (G = 100 Ω, Ig = 1 mA full scale) is to be converted into an AMMETER reading up to 1 A full scale, using a shunt resistance S. The correct value and connection type of S are:

1. S = 100/999 Ω, connected in PARALLEL with G
2. S = 99900 Ω, connected in SERIES with G
3. S = 100/999 Ω, connected in SERIES with G
4. S = 99900 Ω, connected in PARALLEL with G

A galvanometer has resistance G = 100 Ω and gives full-scale deflection for a current Ig = 1 mA. To convert it into a VOLTMETER reading up to 10 V full scale, a resistance Rs must be connected:

1. in parallel with G, with Rs = 100/9 Ω
2. in series with G, with Rs = 9900 Ω
3. in parallel with G, with Rs = 9900 Ω
4. in series with G, with Rs = 100/9 Ω

Resistors R1 = 2 Ω and R2 = 8 Ω are connected to a 10 V ideal battery, first in SERIES, and then (separately) in PARALLEL. The ratio of the total power dissipated in the parallel connection to that in the series connection is:

1. 4
2. 6.25
3. 2
4. 1.6

Two cells, EMF ε1 = 6 V (internal resistance r1 = 2 Ω) and ε2 = 4 V (internal resistance r2 = 2 Ω), are connected in PARALLEL (same polarity together) and this combination drives current through an external resistor R = 9 Ω. The current through cell 2 (the 4 V cell) is:

1. 0.75 A, flowing out of its positive terminal (discharging)
2. 0.25 A, flowing INTO its positive terminal — meaning cell 2 is actually being charged by cell 1, even while the external load simultaneously draws current from the combination
3. 0 A, since both cells have equal internal resistance
4. 2.5 A, flowing out of its positive terminal

A conducting wire is shaped like a truncated cone, with cross-sectional radius decreasing linearly from r1 (thick end) to r2 (thin end, r2 < r1) along its length. A steady current I flows through it. Which statement is TRUE at steady state?

1. The current I is exactly the same at every cross section, while the current density J and drift velocity vd both increase as the wire narrows toward r2
2. The current I decreases as the wire narrows, since resistance increases along the way
3. The current density J is uniform throughout, the same at every cross section, regardless of local area
4. The drift velocity is the same everywhere, since the material and electron density n are unchanged throughout

N identical cells, each of EMF ε and internal resistance r, are arranged in m rows connected in parallel, with n cells connected in series within each row (so N = mn total cells). This battery drives an external resistance R. For the power delivered to R to be MAXIMUM, R must equal:

1. nr/m
2. mr/n
3. nr
4. r/m

A cell of EMF ε and internal resistance r is connected to an external variable resistance R. The value of R for which the power delivered TO R is maximum, and the corresponding maximum power, are:

1. R = 0; Pmax = ε²/r
2. R = r; Pmax = ε²/(4r)
3. R → ∞; Pmax = ε²/r
4. R = 2r; Pmax = ε²/(3r)

A potentiometer is used to find a cell's internal resistance using the standard method: balance length l1 with the cell in open circuit (giving EMF), and balance length l2 with the cell connected across a known resistance R (giving terminal voltage), using r = R(l1−l2)/l2. If the potentiometer's driving (primary) circuit current is now doubled by adjusting the rheostat, the computed value of r:

1. doubles
2. halves
3. remains exactly the same, since l1 and l2 individually change but their combination (l1−l2)/l2 stays invariant
4. becomes undefined, since doubling the driving current invalidates the null method

In a meter bridge experiment, a known resistance R is placed in the LEFT gap and an unknown resistance X in the RIGHT gap, giving a null (balance) point at 40 cm from the left end. If R and X are now interchanged (X in the left gap, R in the right gap), the new balance point, measured from the left end, will be at:

1. 40 cm, unchanged
2. 60 cm
3. 50 cm, always, regardless of R and X
4. Cannot be determined without knowing the individual values of R and X

In a Wheatstone bridge at balance, the battery and the galvanometer are now interchanged — the battery is connected across the diagonal where the galvanometer was, and the galvanometer is connected across the diagonal where the battery was. The bridge:

1. becomes unbalanced, since swapping fundamentally changes the current distribution
2. remains balanced, as a consequence of the reciprocity theorem for linear resistive networks
3. remains balanced only if all four arm resistances happen to be equal
4. cannot be analyzed without knowing the exact resistance values

Twelve equal resistors, each of resistance R, are connected along the twelve edges of a cube. The equivalent resistance between two diagonally opposite corners of the cube (the body diagonal) is:

1. R
2. 5R/6
3. 3R/2
4. 7R/12

A battery of EMF ε and internal resistance r is being CHARGED by an external source, with charging current I flowing INTO its positive terminal. The terminal voltage of the battery during this charging process is:

1. ε − Ir
2. ε + Ir
3. ε, completely unaffected by the charging current
4. Ir − ε

Two resistors, R1 (temperature coefficient of resistance α1, positive) and R2 (temperature coefficient α2), are connected in SERIES such that the combined resistance remains exactly independent of temperature. This requires:

1. α1 = α2, with both positive
2. R1α1 = −R2α2, meaning α2 must have opposite sign to α1 (i.e., R2 must have a negative temperature coefficient)
3. R1 = R2, regardless of the α values
4. This is impossible for any pair of real materials

A conducting wire of length L and cross-sectional area A, made of a given material, is connected across a battery of EMF V (negligible internal resistance), producing a drift velocity vd in its free electrons. The wire is now replaced by another wire of the SAME material and SAME length L, but cross-sectional area 2A, connected to the same battery. The new drift velocity is:

1. vd/2
2. 2vd
3. vd, exactly unchanged
4. 4vd