Current Electricity Test. 2

Red=2, Violet=7, Brown multiplier=10¹, Gold tolerance=5%. This gives 27 × 10¹ = 270 Ω with 5% tolerance.

The 60W bulb has higher resistance (R=V²/P). In series, current is the same, so the bulb with higher resistance dissipates more power (P=I²R) and glows brighter.

For parallel resistors, 1/Req = 1/R₁ + 1/R₂ + 1/R₃ = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 1. Therefore, Req = 1 Ω.

Despite the electric signal traveling near the speed of light, the net forward motion of individual electrons is very slow due to constant collisions.

A voltmeter is connected in parallel, so it must have very high resistance to avoid drawing current from the main circuit and altering the potential difference.

This is the statement of Kirchhoff's Voltage Law (Loop Rule), which is based on the conservation of energy.

A rheostat is a three-terminal variable resistor used to control the amount of current flowing in a circuit.

In electrolytic solutions, the current is carried by the movement of both positive ions (cations) and negative ions (anions).

Current I = Q/t. The total time t = 1 minute + 30 seconds = 60 s + 30 s = 90 s. So, I = 90 C / 90 s = 1.0 A.

Resistance of 25W bulb, R₁ = V²/P = 220²/25 = 1936 Ω. Resistance of 100W bulb, R₂ = 220²/100 = 484 Ω. In series, current I = 440 / (1936+484) = 0.18 A. Voltage across 25W bulb = IR₁ = 0.18 * 1936 ≈ 348 V. Since 348 V is much greater than its rated voltage of 220 V, the 25 W bulb will fuse.

A longer wire leads to a smaller potential gradient (potential drop per unit length), meaning a small voltage corresponds to a larger, more measurable length.

Specific resistance (resistivity) is an intrinsic property of a material that is dependent on its atomic structure and temperature.

Power P = I²R = (2)² × 5 = 20 W. Energy E = Power × time = 20 W × 10 s = 200 J.

For semiconductors, an increase in temperature frees more charge carriers, which increases conductivity and therefore decreases resistance.

Using the shunt formula S = (Ig * G) / (I - Ig). Here, G=95 Ω, Ig = 5 mA = 0.005 A, and I = 5 A. So, S = (0.005 * 95) / (5 - 0.005) = 0.475 / 4.995 ≈ 0.095 Ω.

An ammeter is connected in series, so it must have negligible resistance to avoid altering the current it is intended to measure.

A meter bridge is a practical implementation of the Wheatstone bridge used to find an unknown resistance by achieving a balanced condition.

A semiconductor diode has a non-linear V-I graph, meaning its resistance is not constant and it does not obey Ohm's law.

This is the maximum power transfer theorem, which states that maximum power is delivered to the load when the load resistance equals the source's internal resistance.

When length (L) is doubled, the cross-sectional area (A) is halved (volume is constant). Since R ∝ L/A, the new resistance R' ∝ (2L)/(A/2) = 4(L/A) = 4R.