","comment":{"@type":"Comment","text":"We know that the resistance of a material decreases with increase in conductivity. In such a case, establish a relation between the rate of heat production, the material dimensions and conductivity. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2.64\\,Q$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$3.96\\,Q$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.6\\,Q$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$8.98\\,Q$$","position":2,"answerExplanation":{"@type":"Comment","text":" We know that the rate at which heat is produced is nothing but the power, $$ \\\\ P = \\dfrac{V^2}{R} \\\\ $$ For the given setup,$$V$$ will remain the same for the two since they are connected parallelly, so, $$ \\\\ P \\propto \\dfrac{1}{R} \\\\ $$ We know that $$R = \\dfrac{\\rho l}{A} = \\dfrac{l}{\\sigma A}$$ $$ \\\\ $$ Therefore, $$P \\propto \\dfrac{\\sigma A}{l}$$. $$ \\\\ $$ Given that for constantan: $$ \\\\ \\sigma_c = 9.09 \\times 10^{5}\\,Sm^{-1} \\\\ l_c = 8\\,cm$$ and $$A_c = 2 \\times 0.04 = 0.08\\,cm^2 \\\\ $$ And for nichrome: $$ \\\\ \\sigma_n = 2.04 \\times 10^{6}\\,Sm^{-1} \\\\ l_n = 6\\,cm$$ and $$A_c = 4 \\times 0.06 = 0.24\\,cm^2 \\\\ $$ Therefore, $$ \\\\ \\dfrac{P_n}{P_c} = \\dfrac{R_c}{R_n} \\\\ \\Rightarrow\\dfrac{P_n}{P_c} = \\dfrac{\\sigma_n A_n l_c}{\\sigma_c A_c l_n} \\\\ \\Rightarrow P_n = \\dfrac{\\sigma_n A_n l_c}{\\sigma_c A_c l_n} \\times P_c \\\\ $$ Plugging in the values we get: $$ \\\\ P_{n} = \\dfrac{2.04 \\times 10^6 \\times 0.24 \\times 8}{9.09\\times 10^{5} \\times 0.08 \\times 6} \\times Q \\\\ \\therefore P_n = 8.98\\,Q$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Electric power Quiz 1","text":" Let's presume your monthly home office energy bill is $$Rs.1200$$. You buy a new espresso machine and a minifridge for the office. If the espresso machine operates on a power of $$1000\\,W$$ and is used on an average of $$15\\,min$$ a day, and the minifridge that is kept switched on for $$16\\,hours$$ a day operates at $$250\\,W$$. If your local power station charges $$Rs.5$$ per $$kWh$$, then measure your total monthly electricity bill.","comment":{"@type":"Comment","text":"Determine the individual energy consumption for both the appliances and then find the collective rate of this consumption for the given cost of consumption. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$Rs.1250.5$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$Rs.1221.25$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$Rs.2120.25$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$Rs.1837.5$$","position":1,"answerExplanation":{"@type":"Comment","text":" Given that for the espresso machine: $$ \\\\ $$ $$P = 1000\\,W$$ and $$t = 15min = \\dfrac{15}{60} = 0.25\\,hrs$$ $$ \\\\ $$ The energy consumed by the espresso in a month will be: $$ \\\\ E = P\\times t \\times 30 \\\\ \\Rightarrow E = 1000 \\times 0.25 \\times 30 \\\\ \\Rightarrow E = 7.5\\,kWh \\\\ $$ Similarly, for the minifridge: $$ \\\\ P^{\\prime} = 250\\,W$$ and $$t ^{\\prime}= 16\\,hrs \\\\ $$ The energy consumed by the minifridge in a month will be: $$ \\\\ E^{\\prime} = P^{\\prime} \\times t^{\\prime} \\times 30 \\\\ \\Rightarrow E^{\\prime} = 250 \\times 16 \\times 30 \\\\ \\Rightarrow E^{\\prime} = 120\\,kWh \\\\ $$ The total energy consumed by the espresso machine and the minifridge will be: $$ \\\\ E+E^{\\prime} = 7.5 + 120 \\\\ \\Rightarrow E+E^{\\prime} = 127.5\\,kW \\\\ $$ Given that the power station charges$$Rs.5$$ per$$kWh$$, the cost of energy consumption for the two appliances will be: $$127.5 \\times 5 = Rs. 637.5$$ $$ \\\\ $$ Therefore, the total monthly electricity bill will be: $$Rs.1200 + Rs.637.5 = Rs.1837.5$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Electric power Quiz 1","text":"Suppose that you just bought a glue gun for one of your crafts projects at home. The glue gun has two settings, one for low heat and one for high heat, and the rate at which heat is produced depends on the arrangement of the heating elements either in series or parallel in the glue gun. If one heating element can melt the glue in time $$t_1$$ and the other element can melt the same amount of glue in time $$t_2$$, determine the time taken for the glue gun to melt the glue when the heating elements are connected in series if they take time $$t_{p}$$ to melt the glue when connected in parallel. ","comment":{"@type":"Comment","text":" The amount of heat required to melt the glue will be the same irrespective of which combination of heating elements we use. Plug this into effective resistance expressions to arrive at a suitable result."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{t_{p}}{t_1+ t_2}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{t_{p}}{t_1 t_2}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{t_1+ t_2}{t_{p}}$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{t_1 t_2}{t_{p}}$$","position":3,"answerExplanation":{"@type":"Comment","text":" Let $$V$$ be the supply voltage to the glue gun. Given that $$t_1$$ and $$t_2$$ are the time taken by the first and second heating elements to melt the same amount of glue, the heating elements of the glue gun can be treated as resistive components. Let $$R_1$$ and $$R_2$$ be the resistance of the first and second heating element respectively, and let $$R_s$$ and $$R_p$$ be the effective resistance when they are connected in series and parallel respectively.For a given amount of glue, the amount of heat required to melt it by any heating element will be the same, say, $$Q$$, i.e., $$ \\\\ Q = \\dfrac{V^2}{R_1} \\times t_1 \\\\ \\Rightarrow Q = \\dfrac{V^2}{R_2} \\times t_2 \\\\ \\Rightarrow Q = \\dfrac{V^2}{R_s} \\times t_s \\\\ \\Rightarrow Q = \\dfrac{V^2}{R_p} \\times t_p \\\\ $$ Where $$t_s$$ and $$t_p$$ are the time taken by the glue gun to melt the glue when the heating elements are connected in series and parallel respectively. When the two heating elements are connected in parallel, the net resistance $$R_p$$ can be given as: $$ \\\\ \\dfrac{1}{R_p} = \\dfrac{1}{R_1} + \\dfrac{1}{R_2} \\\\ \\Rightarrow \\dfrac{Q}{V^2 t_p} = \\dfrac{Q}{V^2 t_1} + \\dfrac{Q}{V^2 t_2} \\\\ \\Rightarrow \\dfrac{1}{t_p} = \\dfrac{1}{t_1} + \\dfrac{1}{t_2} = \\dfrac{t_1 +t_2}{t_1 t_2} \\\\ \\Rightarrow t_1+t_2 = \\dfrac{t_1 t_2}{t_p} \\\\ $$ When the two heating elements are connected in series, the net resistance $$R_s$$ can be given as: $$ \\\\ R_s = R_1 + R_2 \\\\ \\Rightarrow \\dfrac{V^2 t_s}{Q} = \\dfrac{V^2 t_1}{Q} +\\dfrac{V^2 t_2}{Q} \\\\ \\Rightarrow t_s = t_1 +t_2 \\\\ $$ But we have that $$ t_1+t_2 = \\dfrac{t_1 t_2}{t_p} \\\\ \\therefore t_s = \\dfrac{t_1 t_2}{t_p}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}