A resistor is made out of a low wire having a length Each end of the wire is att
ID: 1655568 • Letter: A
Question
A resistor is made out of a low wire having a length Each end of the wire is attached to a terminal of a battery having a constant voltage Vo A current I flows through the wire If the wire were cut in half, making two wires of length L/2 and both wire* were attached to Ok battery (the end of both wires attached to one terminal, and the other ends attached to the other terminal). what would be the total current flowing through the two wires? A) I B) 2I C)I D)I/2 Coulomb law f_x=kq_1q_2/r^2, k= 8.988 x 10^9 N-m^2/C^2 almostequalto 9.00 x 10^9 N- m^2/C^2 For point charged particle E= kq/r^2 delta = delta u_e/q for parallel plates delta V= Delta U^e/q= q_+Ed/q= Ed C=Q/V = (1/4 pi k) A/d For 2 charge U1, 2 =k q^1 q^2 /r_1, 2. Capacitance epsilan_0 = 1/4 pi k = 8.85 x 10^-12 c^2/(N-m^2) U_c = 1/2 QV = Q^2/2c = 1/2 CV_2 =q/t, Ohms law V = IR R = p(L/A)Explanation / Answer
we know that resistance of a wire is proportional to its length. Initially, there is one single wire of Length L . The resistance of that wire is R
Current = I=V/R
When the wire is cut into tow halves, the length of each wire becomes L/2 and hence resistance of each wire becomes R/2
Per the setup mentioned in the question, the two wires are coeected in parallel . Equivalent resistance of circuit becomes [(R/2)*(R/2)]/(R/2)+(R/2)] = R/4
Total current flowing through the wires = V/(R/4) = 4V/R = 4(V/R) = 4I
Correct answer is option (A)