Learn maximum power transfer theorem and electrical appliances, calculate equation power delivered to a resistor and dissipated by a resistor. Practice to calculation of electricity bill when more than one electrical appliances are used.
\(P=I^2R\)
\(\Delta V=IR\)
or, \(I=\dfrac {\Delta V}{R}\)
\(P=\left ( \dfrac {\Delta V}{R} \right )^2\;R\)
\(P=\dfrac {(\Delta V)^2}{R} \)
A \(50\) \(\Omega\)
B \(200\) \(\Omega\)
C \(120\) \(\Omega\)
D \(100\) \(\Omega\)
\(P_R=I\;\Delta V_R\)
or, \(P_R=\dfrac {(\Delta V_R)^2}{R}\)
\(E_{th}=P_R\;\Delta t\)
Energy = Watts × seconds
kW \(\rightarrow\)Kilowatt
h \(\rightarrow\) Hour
\(E=P_R\;\Delta t\,\,kWh\)
Here, \(P_R\rightarrow\) power in kW and \(\Delta t\) in hours
Example-
Consider a fan rated at 200 W on a 120 Volt line is used for 5 hrs.
Given :Power = 200 W, Power in kW = 0.2 kW, Time in hrs = 5 hrs
= 0.2 kW × 5 hrs
= 1 kWh
A 400 kW
B 4000 kW
C 4 kW
D 40 kW
Number of units in kWh = (P_{1} + P_{2}) t × 30
Electricity Bill = Number of kWh × Cost / kWh
Electricity Bill = (P_{1} + P_{2}) × t × 30 × Cost / kWh
Electricity Bill = (P_{1}t_{1} + P_{2}t_{2}) × 30 × Cost / kWh
\(P_1=\dfrac {V_1^2}{R_1}\)
or, \(R_1=\dfrac {V_1^2}{P_1}\)
\(P_2=\dfrac {V_2^2}{R_2}\)
or, \(R_2=\dfrac {V_2^2}{P_2}\)
\(I=\dfrac {\mathcal{E}}{\dfrac {V_1^2}{P_1}+\dfrac {V_2^2}{P_2}}\)
\(P_1'=I^2R_1\, \)
where \(P'_1\) is the power dissipated by bulb \(B_1\)
or, \(P_1'=\left ( \dfrac {\mathcal{E}}{R_1+R_2} \right)^2×R_1\)
\(P_2'=I^2R_2\)
where \(P_2'\) is the power dissipated by bulb \(B_2\)
or, \(P_2'=\left ( \dfrac {\mathcal{E}}{R_1+R_2} \right)^2×R_2\)
V_{1} = V_{2} = \(\mathcal{E}\)
Power across B_{1}
\(P_1'=\left (\dfrac {\mathcal{E}}{\dfrac {\mathcal{E}^2}{P_1}+\dfrac {\mathcal{E}^2}{P_2}}\right)^2×\dfrac {\mathcal{E}^2}{P_1}\)
or, \(P_1'=\dfrac {1} {\left (\dfrac {1}{P_1} +\dfrac {1}{P_2}\right)^2}×\dfrac {1}{P_1}\)
or, \(P_1'=\dfrac {1} {\left (\dfrac {P_1+P_2}{P_1P_2} \right)^2}×\dfrac {1}{P_1}\)
or, \(P_1'=\dfrac {P_1^2\;P_2^2} {(P_1+P_2)^2} ×\dfrac {1}{P_1}\)
or, \(P_1'=\dfrac {P_1\;P_2^2} {(P_1+P_2)^2} \)
Similarly,
Power across B_{2}
\(P_2'=\dfrac {P_1^2\;P_2} {(P_1+P_2)^2}\)
A 15 W, 30 W
B 15 W, 20 W
C 20 W, 15 W
D 15 W, 22.5 W
Example-
Electricity Bill = Number of kWh × cost
= 30 × $ 0.10 = $ 3.00
\(I=\dfrac{\text {emf of battery}}{R_{eq}}\)
\(I=\dfrac {\mathcal{E}}{R+r}\)
\(P_R=I^2R\)
\(P_R=\left ( \dfrac {\mathcal{E}}{R+r} \right )^2\cdot R\)
\(\dfrac {dP}{dR}=\dfrac {d}{dR} \left [ \left ( \dfrac {\mathcal{E}}{r+R} \right)^2 \cdot R \right]\)
or, \(\dfrac {dP}{dR}=\mathcal{E}^2\; \dfrac {d}{dR} \left [ \dfrac {R}{(R+r)^2} \right]\)
For power to be maximum, \(\dfrac {dP}{dR}=0\)
\(0=\dfrac {(R+r)^2×1-R×2(R+r)} {(R+r)^4}\)
or,\( 2 R (R + r) = (R + r)^2\)
or, \(R = r\)
A maximum
B less than maximum
C zero
D None of these
P_{1 }< P_{2 }< P_{3} < P_{4}
Since R is inversely proportional to P, so if the voltage rating of all bulbs are same, then the order of resistance is
R_{1 }> R_{2 }> R_{3} > R_{4}
\(P=I^2R\)
or, \(P_{\text { Dissipated}}=I^2R\)
\(P_{\text { D}_1}=I^2R_1\)
\(P_{\text { D}_2}=I^2R_2\)
\(P_{\text { D}_3}=I^2R_3\)
\(P_{\text { D}_4}=I^2R_4\)
\(P\,_{D_1} \propto\;R_1\)
\(P\,_{D_2} \propto\;R_2\)
\(P\,_{D_3} \propto\;R_3\)
\(P\,_{D_4} \propto\;R_4\)
CONCLUSION: The higher the resistance of bulb, the higher the power dissipated and the brighter the bulb glows.
The order of brightness is
B_{1} > B_{2} > B_{3} > B_{4}
A B1 = B2 > B3 > B4 > B5
B B1 < B2 < B3 < B4 < B5
C B1 > B2 > B3 > B4 > B5
D B1 = B2 = B3 = B4 = B5