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This is one of our many quizzes on Electric power. Click on this to access more such competitive quizzes on Current Electricity and EMI.

Electric power Quiz 2

physics

We often leave our phone charger plugged into the outlet, assuming that the charger will not draw any power when it is not being used. This however, is not the case because all plugged in appliances will bleed some energy, which is called standby or phantom electricity that is drawn by appliances in idle mode. If a charger consumes $0.5\,W$ of standby electricity in a period of $360\,mins$, determine the yearly cost of this consumption at a rate of $Rs. 5$ per $kWh$ if the charger is left plugged in all day everyday.    

 Determine the phantom power $\text{(in commercial units)}$ consumed in one year from the given data and arrive at the appropriate cost of consumption. 

  Given that the phantom power consumed by the charger is $10\,mins$ will be $0.5\,W$. $ \\ $ The phantom power consumed by it in one hour will be, $ \\ P = 0.5 \times 6 \\ \Rightarrow P = 3.0\,Wh$ $ \\ $ The phantom power consumed in a year will be, $ \\ P = 3 \times (365 \times 24) \\ \Rightarrow P = 26.28\,kWh \\ $ Therefore, the yearly cost of this consumption at the rate of $Rs. 5$ per $kWh$ will be: $ \\ 26.28 \times 5 = Rs.131.40$

 An electric eel uses high intensity electric discharges to stun or kill prey or defend itself against predators. It has electrolytes in the abdominal region which are essentially biological cells capable of producing electric current by establishing momentary potential differences across their ends.  If the electrolytes have an internal resistance of $315\,\Omega$ due to cellular fluids and minerals and net resistance offered to the high intensity electric pulse by all the electrolytes is about $500\,\Omega$, determine the power output of the eel if it can generate pulses up to $850\,V$. 

 First determine the net resistance offered by the electrolytes following which you can arrive at the power generated.  

 Given that $V = 850\,V$, $r = 315\,\Omega$ and $R = 500\,\Omega$ $ \\ $ The net resistance offered by the electrocytes will be: $ \\ R_{net} = R+r \\ \Rightarrow R_{net}  = 500 +315 \\ \Rightarrow R_{net} = 815\,\Omega \\ $ Therefore, the electric power generated by the eel will be: $ \\ P = \dfrac{V^2}{R_{net}} \\ \Rightarrow P = \dfrac{850^2}{815} \\ \therefore P = 886.5\,W$ 

Determine the power consumed by the 3-band resistor element in the following circuit: $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6ff78d0d-1e85-44b2-9def-4b6628e747c2551817165208105186.png"/> 

Use resistance colour codes to decode the value of resistance offered by the resistor following which you can arrive at the power it consumes. 

 From the figure, we have $V = 120\,V$. $ \\ $ By applying the standard resistance colour-codes to the resistor in the given figure, we arrive at the theoretical resistance of the resistor as follows: $ \\ $ 1st band is equivalent to the first digit: $4 \text{ (Yellow)}$ $ \\ $ 2nd band is equivalent to the second digit: $7 \text{ (Violet)}$  $ \\ $ 3rd band is equivalent to the multiplier: $10^{3} \text{ (Orange)}$ $ \\ $ Thus, the resistance of the resistor will be: $ \\ R = 47 \times 10^{3}\Omega \\ $ Therefore, the power dissipated in the circuit due to the resistor will be: $ \\ P = \dfrac{V^2}{R} \\ \Rightarrow P = \dfrac{120^2}{47 \times 10^3} \\ \therefore P = 0.306\,W$

After learning that copper provides a good level of corrosion resistance, you decide to coat a flat stainless-steel spatula with copper via the process of electroplating as shown below. If the production of copper through this process consumes a power of $96\,kW$, determine the amount of copper liberated per minute in this process, if the electrochemical equivalent of copper is $3.295 \times 10^{-7}\,kgC^{-1}$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d80f2730-be9c-410c-a32d-52b17cca05bf7855184243303803598.png"/> 

First find the current flowing through the circuit following which you can use the definition of the electrochemical equivalent to arrive at the mass of copper that is liberated in this process.  

 Given that power $P = 96\,kW = 96 \times 10^3\,W$, and $V = 120\,V$, $ \\ $ The current flowing through the electrolytic circuit will be: $ \\ I = \dfrac{P}{V} \\ \Rightarrow I = \dfrac{96 \times 10^3}{120} \\ \Rightarrow I = 800\,A \\ $ Now, the electrochemical equivalent $(ECE)$ of copper is given to be  $ \\ z = 3.295 \times 10^{-7}\,kgC^{-1} \\ $ From the units of the ECE we see that, $ \\ z = \dfrac{m}{q} \\ \Rightarrow z = \dfrac{m}{It} \\ $ Therefore, the mass of copper$m$ liberated in time $t = 1\,min = 60\,s$ can be given as: $ \\ m = z \times I \times t \\ \Rightarrow m = 3.295 \times 10^{-7} \times 800 \times 60 \\ \therefore m = 15.8 \times 10^{-3}\,kg$

 Fill in the blanks: $ \\ $ Suppose that a typical lightning bolt of $5 \times 10^{9}\,W$ strikes for a duration of $1.008\,s$, equivalent to __________ $kWh$ of energy. Out of the 1.4 billion lightning strikes per year, only 25% of those strike the ground and can be harnessed. Assuming no capture or transfer losses, the equivalent energy captured is around ___________$kWhyr^{-1}$. $ \\ $ If the world used around $20 \times 10^{12}\,kWh$ in 2020, the number of days through which electrical energy from lightning can power the world is only  about ______ days, which is why though lightning strikes are extremely powerful they cannot be used to harness electricity to commercially power consumer appliances. 

 Begin by converting seconds into hours for the first part. Find the effective number of lightning strikes in the second part to arrive at the appropriate equivalent energy. 

 Given that $P= 5 \times 10^{9}\,W$ and $t = 1.008\,s = \dfrac{1.008}{3600}\,hrs$. $ \\ $ The equivalent energy can be given as: $ \\ E = P \times t \\ \Rightarrow E  = 5 \times 10^9 \times \dfrac{1.008}{3600} \\ \Rightarrow E = 1.4 \times 10^6\,Wh \\ \Rightarrow E = 1400\,kWh \\ $ Therefore, in the context of the question: $ \\ $ “…Suppose that a typical lightning bolt of $5 \times 10^{9}\,W$ strikes for a duration of $1.008s$, equivalent to $1400\,kWh$ of energy.” $ \\ $ Given that only $25\%$ of $1.4 \times 10^9$ lightning strikes per year can be harnessed: $ \\ n = \dfrac{25}{100} \times 1.4 \times 10^9 \\ \Rightarrow n = 35 \times 10^{7}\,yr^{-1} \\ $ The equivalent energy of $n$ number of lightning strikes will be: $ \\ E^{\prime} = E \times n \\ \Rightarrow E^{\prime} = 1400 \times 35 \times 10^{7} \\ \Rightarrow E^{\prime} = 490 \times 10^{9}\,kWhyr^{-1} \\ $ Therefore, in the context of the question: $ \\ $ “…Assuming no capture or transfer losses, the equivalent energy captured is around $490 \times 10^{9}\,kWhyr^{-1}$.” $ \\ $ Given that the world power consumption is $20 \times 10^{12}\,kWh$ in 2020, i.e., $t = 366\,days$, the number of days through which only lightning strikes power the world will be: $ \\ N = \dfrac{490 \times 10^9 \times 366}{20 \times 10^{12}} \\ \therefore N = 8.967 \approx 9\,days \\ $ Therefore, in the context of the question: $ \\ $ “…the number of days through which electrical energy from lightning can power the world is only about 9 days.”