Physics · Ch 2 — Current Electricity

Energy and Power in Electrical Circuits

2.3

Energy and Power in Electrical Circuits

When a battery of voltage V drives current through a circuit, it continuously supplies energy to whatever device is connected. Tracking a small positive charge dQ as it moves once around a simple loop containing the battery and a resistor: moving from point a to b through the battery, the charge GAINS potential energy $dU = V\,dQ$ (at the battery's expense -- its stored chemical energy decreases by exactly this amount); then, moving from c to d through the resistor, the charge LOSES that same energy $dU=V\,dQ$ , dissipated through collisions with the atoms inside the resistor, before returning to point a to repeat the cycle for as long as the battery stays connected.

The rate at which the charge loses this electrical potential energy in the resistor is the electrical power:

P = \dfrac{dU}{dt} = \dfrac{d(V\,dQ)}{dt} = V\dfrac{dQ}{dt} \qquad (2.31)

and since $I=dQ/dt$ , this becomes the compact and central result

P = VI \qquad (2.32)

This gives the power delivered by a battery to ANY electrical system through which current I flows under potential difference V. The SI unit of power is the watt ( $1\ \text{W}=1\ \text{J/s}$ ); commercial light bulbs are routinely labelled with both their power and voltage rating, e.g. 60 W - 220 V. Using Ohm's law, this same power can be re-expressed purely in terms of R and either I or V alone:

P = IV = I(IR) = I^2R \qquad (2.33), \qquad P = IV = V\cdot\dfrac{V}{R} = \dfrac{V^2}{R} \qquad (2.34)

The $I^2R$ form shows that the power a resistor dissipates depends on the SQUARE of the current through it -- doubling the current quadruples the heat dissipated, not merely doubles it (and the same square-law dependence applies to voltage in the $V^2/R$ form). …

Figure 2.15Energy given by the battery

What this figure shows. A single-loop circuit with a battery of voltage V connected to a resistor R, current I flowing around the loop, and four labelled points a, b, c, d marking the corners of the loop -- a and b on the battery side, c and d on the resistor side -- used to track a small positive charge dQ as it gains potential energy $dU=V\,dQ$ moving from a to b through the battery, and loses that same energy to collisions with atoms whil …

Figure 2.16Electrical bulbs with power ratings

What this figure shows. A photograph-style illustration of several household incandescent light bulbs, each stamped with its commercial power-and-voltage rating such as 5W-220V, 30W-220V and 60W-220V, showing how real bulbs are labelled with the same P-and-V numbers used throughout the power formulas in thi …

Misc Example 2.15Comparing brightness of two bulbs in parallel and in series

Worked out. A 30 W bulb and a 60 W bulb are both rated for the same voltage V; the question asks which glows brighter when connected in parallel, which has the greater resistance, and which glows brighter if the two are instead connected in series. (a) In parallel, both bulbs see the same voltage, so $P\propto I$ , meaning the 60 W bulb draws twice the current of the 30 W bulb and is therefore the brighter one in parallel. (b) Using $P=V^2/R$ with the same V for both, $R\propto1/P$ , so the 30 W bulb (lower power) has twice the resistance of the 60 W bulb. (c) In series, both bulbs carry the identical current, so $P=I^2R$ means the bulb with higher resistance dissipates more power -- and since the 30 W bulb has the higher resistance, it is the 30 W bulb, not the 60 W bulb, that glows brighter when the two are wired in series. The overall lesson: a bulb's power rating alone does not decide which glows brigh …

Misc Example 2.16Which of two series bulbs will fuse on an oversupply

Worked out. A 20 W - 220 V bulb and a 100 W - 220 V bulb are connected in series to a 440 V supply, and the question asks which bulb will fuse (burn out). Each bulb's own resistance is found from $R = V_{rated}^2/P_{rated}$ : the 20 W bulb has $R_1 = 220^2/20 = 2420\ \Omega$ , and the 100 W bulb has $R_2 = 220^2/100 = 484\ \Omega$ . In series, the same current flows through both: $I = 440/(R_1+R_2) = 440/2904 \approx 0.151$ A. The voltage drop across the 20 W bulb is then $V_1=IR_1\approx366.6$ V, and across the 100 W bulb is $V_2=IR_2\approx73.3$ V. Since $V_1$ (about 367 V) far exceeds the 20 W bulb's 220 V rating, while $V_2$ stays well under the 100 W bulb's rating, the 20 W bulb -- counter-intuitively the LOWER-wattage bulb -- is the one that fuses, because in series it is the higher- …

Misc ~note-tneb-billingWhat the electricity board actually charges for

Worked out. A short practical note: the Tamil Nadu Electricity Board charges customers for the amount of ENERGY consumed, not for power -- power (in watts) only tells you the rate of energy use at an instant, while the electricity bill is based on total energy (power multiplied by the time each appliance runs), measured commercially in kilowatt-hours (units). …

Start your 14-day free trial to unlock the full solution →

← Previous · 2.2.4Temperature Dependence of Resistivity Next · 2.4 →Electric Cells and Batteries