Current Electricity Test. 5

The slope is ΔI/ΔV. Since resistance R = V/I, the slope is 1/R, which is defined as conductance (G).

Using the formula E = I(R + r). We have 6 = 1.0(R + 0.5). So, 6 = R + 0.5, which gives R = 5.5 Ω.

The maximum potential drop that can be measured by the potentiometer is the EMF of its driver cell. Therefore, the driver EMF must be greater than the unknown EMF to find a balance point.

The low-resistance shunt bypasses or 'shunts' the majority of the current around the sensitive galvanometer, allowing the instrument to measure a much larger total current.

Relaxation time is the average time between collisions. Lowering the temperature reduces the thermal vibrations of the metal lattice, leading to fewer collisions and thus a longer average time between them.

From the equation I = nAv_d, if current (I) is constant and area (A) decreases, the drift velocity (v_d) must increase to maintain the equality.

Semiconductors like silicon and germanium have electrical properties intermediate to those of good conductors and good insulators.

Using R_T = R₀(1 + αΔT), where ΔT = T_final - T_initial. 20 = 5(1 + 0.004(T - 20)). This simplifies to 4 = 1 + 0.004(T - 20), so 3 = 0.004(T-20). T-20 = 3/0.004 = 750. Therefore, T = 750 + 20 = 770°C.

Current is charge/time (A = C/s). Therefore, charge = current × time. Ampere-hour is a unit of current multiplied by time, making it a unit of electric charge, commonly used for battery capacity.

Using E = I(R + r), we have 1.5 = 2(0.5 + r). So, 0.75 = 0.5 + r, which gives the internal resistance r = 0.25 Ω.

A resistor is a passive component that only dissipates electrical energy as heat. Sources of EMF (like cells, generators) convert other forms of energy into electrical energy.

The meter bridge follows the Wheatstone principle: R₁/R₂ = L₁/L₂. So, 6/R_unknown = 40/(100-40) = 40/60. R_unknown = 6 × (60/40) = 9 Ω.

An electric fuse contains a wire with a low melting point that heats up and melts (breaks the circuit) when the current exceeds a safe level, based on Joule's law of heating.

Using the power formula P = V²/R, we can solve for resistance: R = V²/P = (6V)² / 12W = 36/12 = 3 Ω.

Devices like semiconductor diodes, transistors, and vacuum tubes are non-ohmic because their current-voltage relationship is not linear.