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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 94 "restarts are always good to clear all the
 memory from previous work\nThen define the resistors;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "R1:=10:\nR2:=6:\nR3:=30:\ni:=3:" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "R2 and R3 are in parallel, so we \+
calculate the equivalent resistance Req = 1/(1/R1 + 1/R2)" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Req:=1/(1/R2 + 1/R3);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 66 "Now we work out the equilivent for the en
tire circuit; ie Req + R1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
17 "R[total]:=Req+R1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 190 "Now we \+
can work out the voltage across the battery (since its the same as wha
ts across the total resistance) using the current and the equivalent t
otal Resistance using V=IR\nWe have I and R," }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 14 "V:=i*R[total];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 113 "Now the current through each resistor is defined by the voltag
e accross that resistor divided by that resistance." }}{PARA 0 "" 0 "
" {TEXT -1 289 "For the first resistor, all the current from the batte
ry passed through that resistor, so the current is 3A.\nFor the second
 resistors, we need to find the voltage across them, so we take into a
ccount the voltage drop from the first resistor.\n\ni*R1 is the voltag
e across the first resistor." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 18 "V[second]:=V-i*R1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Then \+
the respective currents are simply:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "i[1]:=3;\ni[2]:=V[second]/R2;\ni[3]:=V[second]/R3;" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "14" 0 }
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