Three resistors, R1= 29ohms , R2 = 78ohms , and R3=R , are connected in parallel
ID: 1687813 • Letter: T
Question
Three resistors, R1= 29ohms , R2 = 78ohms , and R3=R , are connected in parallel with a 12 battery. The total current flowing through the battery is .88A.A. Find the current through each resistor. B. If the total current in the battery had been greater than .88A, would your answer to part A have been larger or smaller?
** I know the formula to solve is 1/R= 1/R1+1/R2+1/R3
I enter 1/29+1/78+1/38 into my calculator, and get .0736, but that's not the answer. A. Find the current through each resistor. B. If the total current in the battery had been greater than .88A, would your answer to part A have been larger or smaller?
Explanation / Answer
resistors R1= 29 Ohms , R2 = 78 Ohms and R3=38 Ohms current I =0.88 A voltage V =12 V A) current through resister R_1 is I_1 =V/R_1 =12 V/29 Ohms =0.413 A current through resister R_2 is I_2 =V/R_2 =12 V/78 Ohms =0.153 A current through resister R_3 is I_3 =V/R_3 =12 V/38 Ohms =0.315 A B) Three resistors are connected in parallel with a 12 V battery resistance 1/R =1/R_1+1/R_2+1/R_3 1/R =0.0735 R =13.60 Ohms Then the total voltage in the circuit is V = IR I =V/R =0.882 A Here the resistances R_1, R_2 and R_3 doesn't chage, so the value of resultant resistance R doesnot change Then if the current (I) is increases (greater than previous current value) then the value of voltage (V) also increases [since V = IR] Then I_1 = V / R_1 Here R_1 not changing so I_1 is proportional to V, so I_1 increases. And I_2 = V / R_2 Here R_2 not changing so I_2 is proportional to V, so I_2 increases. And I_3 = V / R_3 Here R_3 not changing so I_3 is proportional to V, so I_3 increases.
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